Existence of transverse homotopy between knots in a 3-manifold












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I have a 3-manifold $Sigma$ and two homotopic embedded knots $K_{0}(t): S^{1} to Sigma$ and $K_{1}(t): S^{1} to Sigma$. I wish to refine the homotopy between them to a "transverse homotopy" i.e, some $H(s,t)$ such that




  • $s in {s_{0},ldots,s_{n}} Rightarrow H(s,t)$ is a singular knot with one crossing i.e, there are only two points $t_{1},t_{2}in[0,1]$ such that $H(s,t_{1})=H(s,t_{2})$.


  • $s in [0,1]backslash {s_{0},ldots,s_{n}} Rightarrow H(s,t)$ is an embedded knot.



In definition 2.1 of Type 1 knot invariants in 3-manifolds (bottom of page 5) the authors state that a transverse homotopy always exists between any two homotopic knots.



I'd like to know how to prove this statement.










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    $begingroup$


    I have a 3-manifold $Sigma$ and two homotopic embedded knots $K_{0}(t): S^{1} to Sigma$ and $K_{1}(t): S^{1} to Sigma$. I wish to refine the homotopy between them to a "transverse homotopy" i.e, some $H(s,t)$ such that




    • $s in {s_{0},ldots,s_{n}} Rightarrow H(s,t)$ is a singular knot with one crossing i.e, there are only two points $t_{1},t_{2}in[0,1]$ such that $H(s,t_{1})=H(s,t_{2})$.


    • $s in [0,1]backslash {s_{0},ldots,s_{n}} Rightarrow H(s,t)$ is an embedded knot.



    In definition 2.1 of Type 1 knot invariants in 3-manifolds (bottom of page 5) the authors state that a transverse homotopy always exists between any two homotopic knots.



    I'd like to know how to prove this statement.










    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      I have a 3-manifold $Sigma$ and two homotopic embedded knots $K_{0}(t): S^{1} to Sigma$ and $K_{1}(t): S^{1} to Sigma$. I wish to refine the homotopy between them to a "transverse homotopy" i.e, some $H(s,t)$ such that




      • $s in {s_{0},ldots,s_{n}} Rightarrow H(s,t)$ is a singular knot with one crossing i.e, there are only two points $t_{1},t_{2}in[0,1]$ such that $H(s,t_{1})=H(s,t_{2})$.


      • $s in [0,1]backslash {s_{0},ldots,s_{n}} Rightarrow H(s,t)$ is an embedded knot.



      In definition 2.1 of Type 1 knot invariants in 3-manifolds (bottom of page 5) the authors state that a transverse homotopy always exists between any two homotopic knots.



      I'd like to know how to prove this statement.










      share|cite|improve this question









      $endgroup$




      I have a 3-manifold $Sigma$ and two homotopic embedded knots $K_{0}(t): S^{1} to Sigma$ and $K_{1}(t): S^{1} to Sigma$. I wish to refine the homotopy between them to a "transverse homotopy" i.e, some $H(s,t)$ such that




      • $s in {s_{0},ldots,s_{n}} Rightarrow H(s,t)$ is a singular knot with one crossing i.e, there are only two points $t_{1},t_{2}in[0,1]$ such that $H(s,t_{1})=H(s,t_{2})$.


      • $s in [0,1]backslash {s_{0},ldots,s_{n}} Rightarrow H(s,t)$ is an embedded knot.



      In definition 2.1 of Type 1 knot invariants in 3-manifolds (bottom of page 5) the authors state that a transverse homotopy always exists between any two homotopic knots.



      I'd like to know how to prove this statement.







      algebraic-topology manifolds knot-theory knot-invariants






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      asked Dec 6 '18 at 17:44









      Mike KissMike Kiss

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          Suppose $h : S^1 times I to Sigma$ is a smooth homotopy between the knots $K_0 = h(-, 0)$ and $K_1 = h(-, 1)$. Consider the movie $f : S^1 times I to Sigma times I$ of the homotopy, defined by $f(x, t) = (h(x, t), t)$.



          I claim that $f$ is homotopic, relative to it's boundary, to a map $g$ which is an immersion with transverse double points, i.e., $g$ is an embedding away from finitely many points $x_1, y_1, cdots, x_n, y_n in S^1 times I$ such that $g(x_k) = g(y_k) = p_k$ for $1 leq k leq n$ and $text{im} ,dg_{x_k} pitchfork text{im} , dg_{y_k}$. This follows from a self-transversality theorem that I have stated and given a sketch of proof below.



          As a matter of fact $g$ can be chosen to be $C^1$-close to $f$ as well. $f$ was completely horizontal to $Sigma$, i.e., if we denote $f_t = f|S^1 times {t}$ to be the restriction on the $t$-slice, $f_t' subset (dpi)^*TSigma$ where $pi : Sigma times I to Sigma$ is the projection. Therefore $g_t (=g|S^1 times {t})$ must be $varepsilon$-horizontal to $Sigma$, and in particular, $g_t' notin ker dpi$. This would mean $pi circ g_t$ is an immersion for all $t in I$.



          Note that this is almost enough for what you want: Take $H = pi_Sigma circ g$ to be your new homotopy. This is a homotopy of $K_0$ to $K_1$ through immersions, and the only difficulty is that $H_t$ might stay on a singular knot for an interval's worth of time $t in J subset I$, whereas you want it to stay there for only finitely many points in $I$. Locally near a crossing, $(H_t)_{t in I}$ would look like a pair of undercrossing-overcrossing which gradually grows closer, stays in the position of "$mathsf{X}$" for the interval $t in J$ and then grows apart and becomes a pair of overcrossing-undercrossing. We modify $H$ near those points so that it only stays like an "$mathsf{X}$" for a point $t = t_0 in J$. Then further modify $H$ so that double points are introduced one at a time, i.e, $H_t$ has at most one double point.





          Suppose $M$ is a compact $n$-manifold and $N$ is a compact $2n$-manifold. Let $C^infty(M, N)$ be the space of smooth maps and $text{Imm}(M, N)$ be the subspace of such maps which are immersions.



          Call a map $f in text{Imm}(M, N)$ to be an immersion with clean double points if whenever $x, y in M$, and $f(x) = f(y) = p$, there exists charts $U, V$ around $x, y$ respectively such that $f|_U, f|_V$ are embeddings, $f(U)$ intersects $f(V)$ transversely, and $f(U) cap f(V) = {p}$. Denote the subspace of such maps as $text{Imm}_{pitchfork}(M, N)$



          Theorem (Self-transversality of maps): $text{Imm}_pitchfork(M^n, N^{2n})$ is dense in $C^infty(M^n, N^{2n})$.



          Let $J^1(M, N)$ be the space of $1$-jets of maps $M to N$ and $mathscr{S} subset J^1(M, N)$ be the subset of $1$-jets of non-immersions. The space of $1$-jets is an affine bundle $M_{2n times n}(Bbb R) to J^1(M, N) to M times N$ with fiber keeping track of the formal derivative component. Let $Sigma_{< n} subset M_{2n times n}(Bbb R)$ be the stratified subset of matrices of rank strictly less than $n$. As $mathscr{S}$ fibers over $M times N$ with fiber $Sigma_{< n}$, it is also a stratified subset of $J^1(M, N)$.



          Given any map $f in C^infty(M, N)$, consider it's 1-jet prolongation $j^1 f : M to J^1(M, N)$. By Thom transversality theorem, we can homotope $f$ by a $C^1$-small homotopy to $g in C^infty(M, N)$ such that $j^1 g pitchfork mathscr{S}$. But observe that $dim J^1(M, N) = n(2n+3)$, so $text{codim}, j^1g(M) = dim J^1(M, N)$ $- dim M$ $=$ $2n(n+1)$ whereas $text{codim}, mathscr{S}$ $=$ $text{codim}_{M_{2n times n}(Bbb R)} Sigma_{< n}$ $=$ $n+1$, so $text{codim}, j^1 g(M)$ $+$ $text{codim}, mathscr{S}$ $>$ $dim J^1(M, N)$, forcing $j^1 g$ to be disjoint from $mathscr{S}$. This implies $j^1 g$ has rank $n$ everywhere, i.e., $g$ is an immersion.



          This implies $text{Imm}(M, N)$ is dense in $C^infty(M, N)$. Now for any $f in text{Imm}(M, N)$ consider the map $F : M times M to N times N$, $F(x, y) = (f(x), f(y))$. By a $C^1$-small homotopy we can modify $F$ to $G$ so that $G$ is transverse to the diagonal $Delta_N subset N times N$. Let $G|Delta_M$ be the restriction to the diagonal of $M times M$. As $G$ is $C^1$-close to $F$, image of $G|Delta_M$ fits inside an $epsilon$-neighborhood of $Delta_N subset N times N$, and by $epsilon$-neighborhood theorem we have a normal projection map to $Delta_N$, with which we compose to get a map $g : M to N$. $g times g$ is $C^1$-close to $G$ near $Delta_M$, and since $G pitchfork Delta_N$, by stability of transverse maps, $(g times g) pitchfork Delta_M$ should hold as well. Therefore $g in text{Imm}_pitchfork(M, N)$.



          This implies we have a tower of successively dense subspaces $$text{Imm}_pitchfork(M, N) subset text{Imm}(M, N) subset C^infty(M, N)$$
          In particular, $text{Imm}_pitchfork(M, N)$ is dense in $C^infty(M, N)$, as required. (Note that for the above we need a relative-to-boundary version of this, but once $f$ is already an embedding restricted to (a collar neighborhood of) the boundary, we can do all the required homotopies fixing the boundary.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So does this result only hold if $Sigma$ is compact? Or are there analogues of the Thom transversality theorem for maps from compact n-manifolds to general 2n-manifolds?
            $endgroup$
            – Mike Kiss
            Dec 7 '18 at 14:26






          • 1




            $begingroup$
            @MikeKiss Compactness of the codomain is not necessary for transversality results to hold, no. Also note that my argument on the first bit is incomplete; I proposed a possible way to fix it. I'll think about it and see if it works.
            $endgroup$
            – Balarka Sen
            Dec 7 '18 at 15:57











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          $begingroup$

          Suppose $h : S^1 times I to Sigma$ is a smooth homotopy between the knots $K_0 = h(-, 0)$ and $K_1 = h(-, 1)$. Consider the movie $f : S^1 times I to Sigma times I$ of the homotopy, defined by $f(x, t) = (h(x, t), t)$.



          I claim that $f$ is homotopic, relative to it's boundary, to a map $g$ which is an immersion with transverse double points, i.e., $g$ is an embedding away from finitely many points $x_1, y_1, cdots, x_n, y_n in S^1 times I$ such that $g(x_k) = g(y_k) = p_k$ for $1 leq k leq n$ and $text{im} ,dg_{x_k} pitchfork text{im} , dg_{y_k}$. This follows from a self-transversality theorem that I have stated and given a sketch of proof below.



          As a matter of fact $g$ can be chosen to be $C^1$-close to $f$ as well. $f$ was completely horizontal to $Sigma$, i.e., if we denote $f_t = f|S^1 times {t}$ to be the restriction on the $t$-slice, $f_t' subset (dpi)^*TSigma$ where $pi : Sigma times I to Sigma$ is the projection. Therefore $g_t (=g|S^1 times {t})$ must be $varepsilon$-horizontal to $Sigma$, and in particular, $g_t' notin ker dpi$. This would mean $pi circ g_t$ is an immersion for all $t in I$.



          Note that this is almost enough for what you want: Take $H = pi_Sigma circ g$ to be your new homotopy. This is a homotopy of $K_0$ to $K_1$ through immersions, and the only difficulty is that $H_t$ might stay on a singular knot for an interval's worth of time $t in J subset I$, whereas you want it to stay there for only finitely many points in $I$. Locally near a crossing, $(H_t)_{t in I}$ would look like a pair of undercrossing-overcrossing which gradually grows closer, stays in the position of "$mathsf{X}$" for the interval $t in J$ and then grows apart and becomes a pair of overcrossing-undercrossing. We modify $H$ near those points so that it only stays like an "$mathsf{X}$" for a point $t = t_0 in J$. Then further modify $H$ so that double points are introduced one at a time, i.e, $H_t$ has at most one double point.





          Suppose $M$ is a compact $n$-manifold and $N$ is a compact $2n$-manifold. Let $C^infty(M, N)$ be the space of smooth maps and $text{Imm}(M, N)$ be the subspace of such maps which are immersions.



          Call a map $f in text{Imm}(M, N)$ to be an immersion with clean double points if whenever $x, y in M$, and $f(x) = f(y) = p$, there exists charts $U, V$ around $x, y$ respectively such that $f|_U, f|_V$ are embeddings, $f(U)$ intersects $f(V)$ transversely, and $f(U) cap f(V) = {p}$. Denote the subspace of such maps as $text{Imm}_{pitchfork}(M, N)$



          Theorem (Self-transversality of maps): $text{Imm}_pitchfork(M^n, N^{2n})$ is dense in $C^infty(M^n, N^{2n})$.



          Let $J^1(M, N)$ be the space of $1$-jets of maps $M to N$ and $mathscr{S} subset J^1(M, N)$ be the subset of $1$-jets of non-immersions. The space of $1$-jets is an affine bundle $M_{2n times n}(Bbb R) to J^1(M, N) to M times N$ with fiber keeping track of the formal derivative component. Let $Sigma_{< n} subset M_{2n times n}(Bbb R)$ be the stratified subset of matrices of rank strictly less than $n$. As $mathscr{S}$ fibers over $M times N$ with fiber $Sigma_{< n}$, it is also a stratified subset of $J^1(M, N)$.



          Given any map $f in C^infty(M, N)$, consider it's 1-jet prolongation $j^1 f : M to J^1(M, N)$. By Thom transversality theorem, we can homotope $f$ by a $C^1$-small homotopy to $g in C^infty(M, N)$ such that $j^1 g pitchfork mathscr{S}$. But observe that $dim J^1(M, N) = n(2n+3)$, so $text{codim}, j^1g(M) = dim J^1(M, N)$ $- dim M$ $=$ $2n(n+1)$ whereas $text{codim}, mathscr{S}$ $=$ $text{codim}_{M_{2n times n}(Bbb R)} Sigma_{< n}$ $=$ $n+1$, so $text{codim}, j^1 g(M)$ $+$ $text{codim}, mathscr{S}$ $>$ $dim J^1(M, N)$, forcing $j^1 g$ to be disjoint from $mathscr{S}$. This implies $j^1 g$ has rank $n$ everywhere, i.e., $g$ is an immersion.



          This implies $text{Imm}(M, N)$ is dense in $C^infty(M, N)$. Now for any $f in text{Imm}(M, N)$ consider the map $F : M times M to N times N$, $F(x, y) = (f(x), f(y))$. By a $C^1$-small homotopy we can modify $F$ to $G$ so that $G$ is transverse to the diagonal $Delta_N subset N times N$. Let $G|Delta_M$ be the restriction to the diagonal of $M times M$. As $G$ is $C^1$-close to $F$, image of $G|Delta_M$ fits inside an $epsilon$-neighborhood of $Delta_N subset N times N$, and by $epsilon$-neighborhood theorem we have a normal projection map to $Delta_N$, with which we compose to get a map $g : M to N$. $g times g$ is $C^1$-close to $G$ near $Delta_M$, and since $G pitchfork Delta_N$, by stability of transverse maps, $(g times g) pitchfork Delta_M$ should hold as well. Therefore $g in text{Imm}_pitchfork(M, N)$.



          This implies we have a tower of successively dense subspaces $$text{Imm}_pitchfork(M, N) subset text{Imm}(M, N) subset C^infty(M, N)$$
          In particular, $text{Imm}_pitchfork(M, N)$ is dense in $C^infty(M, N)$, as required. (Note that for the above we need a relative-to-boundary version of this, but once $f$ is already an embedding restricted to (a collar neighborhood of) the boundary, we can do all the required homotopies fixing the boundary.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So does this result only hold if $Sigma$ is compact? Or are there analogues of the Thom transversality theorem for maps from compact n-manifolds to general 2n-manifolds?
            $endgroup$
            – Mike Kiss
            Dec 7 '18 at 14:26






          • 1




            $begingroup$
            @MikeKiss Compactness of the codomain is not necessary for transversality results to hold, no. Also note that my argument on the first bit is incomplete; I proposed a possible way to fix it. I'll think about it and see if it works.
            $endgroup$
            – Balarka Sen
            Dec 7 '18 at 15:57
















          4












          $begingroup$

          Suppose $h : S^1 times I to Sigma$ is a smooth homotopy between the knots $K_0 = h(-, 0)$ and $K_1 = h(-, 1)$. Consider the movie $f : S^1 times I to Sigma times I$ of the homotopy, defined by $f(x, t) = (h(x, t), t)$.



          I claim that $f$ is homotopic, relative to it's boundary, to a map $g$ which is an immersion with transverse double points, i.e., $g$ is an embedding away from finitely many points $x_1, y_1, cdots, x_n, y_n in S^1 times I$ such that $g(x_k) = g(y_k) = p_k$ for $1 leq k leq n$ and $text{im} ,dg_{x_k} pitchfork text{im} , dg_{y_k}$. This follows from a self-transversality theorem that I have stated and given a sketch of proof below.



          As a matter of fact $g$ can be chosen to be $C^1$-close to $f$ as well. $f$ was completely horizontal to $Sigma$, i.e., if we denote $f_t = f|S^1 times {t}$ to be the restriction on the $t$-slice, $f_t' subset (dpi)^*TSigma$ where $pi : Sigma times I to Sigma$ is the projection. Therefore $g_t (=g|S^1 times {t})$ must be $varepsilon$-horizontal to $Sigma$, and in particular, $g_t' notin ker dpi$. This would mean $pi circ g_t$ is an immersion for all $t in I$.



          Note that this is almost enough for what you want: Take $H = pi_Sigma circ g$ to be your new homotopy. This is a homotopy of $K_0$ to $K_1$ through immersions, and the only difficulty is that $H_t$ might stay on a singular knot for an interval's worth of time $t in J subset I$, whereas you want it to stay there for only finitely many points in $I$. Locally near a crossing, $(H_t)_{t in I}$ would look like a pair of undercrossing-overcrossing which gradually grows closer, stays in the position of "$mathsf{X}$" for the interval $t in J$ and then grows apart and becomes a pair of overcrossing-undercrossing. We modify $H$ near those points so that it only stays like an "$mathsf{X}$" for a point $t = t_0 in J$. Then further modify $H$ so that double points are introduced one at a time, i.e, $H_t$ has at most one double point.





          Suppose $M$ is a compact $n$-manifold and $N$ is a compact $2n$-manifold. Let $C^infty(M, N)$ be the space of smooth maps and $text{Imm}(M, N)$ be the subspace of such maps which are immersions.



          Call a map $f in text{Imm}(M, N)$ to be an immersion with clean double points if whenever $x, y in M$, and $f(x) = f(y) = p$, there exists charts $U, V$ around $x, y$ respectively such that $f|_U, f|_V$ are embeddings, $f(U)$ intersects $f(V)$ transversely, and $f(U) cap f(V) = {p}$. Denote the subspace of such maps as $text{Imm}_{pitchfork}(M, N)$



          Theorem (Self-transversality of maps): $text{Imm}_pitchfork(M^n, N^{2n})$ is dense in $C^infty(M^n, N^{2n})$.



          Let $J^1(M, N)$ be the space of $1$-jets of maps $M to N$ and $mathscr{S} subset J^1(M, N)$ be the subset of $1$-jets of non-immersions. The space of $1$-jets is an affine bundle $M_{2n times n}(Bbb R) to J^1(M, N) to M times N$ with fiber keeping track of the formal derivative component. Let $Sigma_{< n} subset M_{2n times n}(Bbb R)$ be the stratified subset of matrices of rank strictly less than $n$. As $mathscr{S}$ fibers over $M times N$ with fiber $Sigma_{< n}$, it is also a stratified subset of $J^1(M, N)$.



          Given any map $f in C^infty(M, N)$, consider it's 1-jet prolongation $j^1 f : M to J^1(M, N)$. By Thom transversality theorem, we can homotope $f$ by a $C^1$-small homotopy to $g in C^infty(M, N)$ such that $j^1 g pitchfork mathscr{S}$. But observe that $dim J^1(M, N) = n(2n+3)$, so $text{codim}, j^1g(M) = dim J^1(M, N)$ $- dim M$ $=$ $2n(n+1)$ whereas $text{codim}, mathscr{S}$ $=$ $text{codim}_{M_{2n times n}(Bbb R)} Sigma_{< n}$ $=$ $n+1$, so $text{codim}, j^1 g(M)$ $+$ $text{codim}, mathscr{S}$ $>$ $dim J^1(M, N)$, forcing $j^1 g$ to be disjoint from $mathscr{S}$. This implies $j^1 g$ has rank $n$ everywhere, i.e., $g$ is an immersion.



          This implies $text{Imm}(M, N)$ is dense in $C^infty(M, N)$. Now for any $f in text{Imm}(M, N)$ consider the map $F : M times M to N times N$, $F(x, y) = (f(x), f(y))$. By a $C^1$-small homotopy we can modify $F$ to $G$ so that $G$ is transverse to the diagonal $Delta_N subset N times N$. Let $G|Delta_M$ be the restriction to the diagonal of $M times M$. As $G$ is $C^1$-close to $F$, image of $G|Delta_M$ fits inside an $epsilon$-neighborhood of $Delta_N subset N times N$, and by $epsilon$-neighborhood theorem we have a normal projection map to $Delta_N$, with which we compose to get a map $g : M to N$. $g times g$ is $C^1$-close to $G$ near $Delta_M$, and since $G pitchfork Delta_N$, by stability of transverse maps, $(g times g) pitchfork Delta_M$ should hold as well. Therefore $g in text{Imm}_pitchfork(M, N)$.



          This implies we have a tower of successively dense subspaces $$text{Imm}_pitchfork(M, N) subset text{Imm}(M, N) subset C^infty(M, N)$$
          In particular, $text{Imm}_pitchfork(M, N)$ is dense in $C^infty(M, N)$, as required. (Note that for the above we need a relative-to-boundary version of this, but once $f$ is already an embedding restricted to (a collar neighborhood of) the boundary, we can do all the required homotopies fixing the boundary.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So does this result only hold if $Sigma$ is compact? Or are there analogues of the Thom transversality theorem for maps from compact n-manifolds to general 2n-manifolds?
            $endgroup$
            – Mike Kiss
            Dec 7 '18 at 14:26






          • 1




            $begingroup$
            @MikeKiss Compactness of the codomain is not necessary for transversality results to hold, no. Also note that my argument on the first bit is incomplete; I proposed a possible way to fix it. I'll think about it and see if it works.
            $endgroup$
            – Balarka Sen
            Dec 7 '18 at 15:57














          4












          4








          4





          $begingroup$

          Suppose $h : S^1 times I to Sigma$ is a smooth homotopy between the knots $K_0 = h(-, 0)$ and $K_1 = h(-, 1)$. Consider the movie $f : S^1 times I to Sigma times I$ of the homotopy, defined by $f(x, t) = (h(x, t), t)$.



          I claim that $f$ is homotopic, relative to it's boundary, to a map $g$ which is an immersion with transverse double points, i.e., $g$ is an embedding away from finitely many points $x_1, y_1, cdots, x_n, y_n in S^1 times I$ such that $g(x_k) = g(y_k) = p_k$ for $1 leq k leq n$ and $text{im} ,dg_{x_k} pitchfork text{im} , dg_{y_k}$. This follows from a self-transversality theorem that I have stated and given a sketch of proof below.



          As a matter of fact $g$ can be chosen to be $C^1$-close to $f$ as well. $f$ was completely horizontal to $Sigma$, i.e., if we denote $f_t = f|S^1 times {t}$ to be the restriction on the $t$-slice, $f_t' subset (dpi)^*TSigma$ where $pi : Sigma times I to Sigma$ is the projection. Therefore $g_t (=g|S^1 times {t})$ must be $varepsilon$-horizontal to $Sigma$, and in particular, $g_t' notin ker dpi$. This would mean $pi circ g_t$ is an immersion for all $t in I$.



          Note that this is almost enough for what you want: Take $H = pi_Sigma circ g$ to be your new homotopy. This is a homotopy of $K_0$ to $K_1$ through immersions, and the only difficulty is that $H_t$ might stay on a singular knot for an interval's worth of time $t in J subset I$, whereas you want it to stay there for only finitely many points in $I$. Locally near a crossing, $(H_t)_{t in I}$ would look like a pair of undercrossing-overcrossing which gradually grows closer, stays in the position of "$mathsf{X}$" for the interval $t in J$ and then grows apart and becomes a pair of overcrossing-undercrossing. We modify $H$ near those points so that it only stays like an "$mathsf{X}$" for a point $t = t_0 in J$. Then further modify $H$ so that double points are introduced one at a time, i.e, $H_t$ has at most one double point.





          Suppose $M$ is a compact $n$-manifold and $N$ is a compact $2n$-manifold. Let $C^infty(M, N)$ be the space of smooth maps and $text{Imm}(M, N)$ be the subspace of such maps which are immersions.



          Call a map $f in text{Imm}(M, N)$ to be an immersion with clean double points if whenever $x, y in M$, and $f(x) = f(y) = p$, there exists charts $U, V$ around $x, y$ respectively such that $f|_U, f|_V$ are embeddings, $f(U)$ intersects $f(V)$ transversely, and $f(U) cap f(V) = {p}$. Denote the subspace of such maps as $text{Imm}_{pitchfork}(M, N)$



          Theorem (Self-transversality of maps): $text{Imm}_pitchfork(M^n, N^{2n})$ is dense in $C^infty(M^n, N^{2n})$.



          Let $J^1(M, N)$ be the space of $1$-jets of maps $M to N$ and $mathscr{S} subset J^1(M, N)$ be the subset of $1$-jets of non-immersions. The space of $1$-jets is an affine bundle $M_{2n times n}(Bbb R) to J^1(M, N) to M times N$ with fiber keeping track of the formal derivative component. Let $Sigma_{< n} subset M_{2n times n}(Bbb R)$ be the stratified subset of matrices of rank strictly less than $n$. As $mathscr{S}$ fibers over $M times N$ with fiber $Sigma_{< n}$, it is also a stratified subset of $J^1(M, N)$.



          Given any map $f in C^infty(M, N)$, consider it's 1-jet prolongation $j^1 f : M to J^1(M, N)$. By Thom transversality theorem, we can homotope $f$ by a $C^1$-small homotopy to $g in C^infty(M, N)$ such that $j^1 g pitchfork mathscr{S}$. But observe that $dim J^1(M, N) = n(2n+3)$, so $text{codim}, j^1g(M) = dim J^1(M, N)$ $- dim M$ $=$ $2n(n+1)$ whereas $text{codim}, mathscr{S}$ $=$ $text{codim}_{M_{2n times n}(Bbb R)} Sigma_{< n}$ $=$ $n+1$, so $text{codim}, j^1 g(M)$ $+$ $text{codim}, mathscr{S}$ $>$ $dim J^1(M, N)$, forcing $j^1 g$ to be disjoint from $mathscr{S}$. This implies $j^1 g$ has rank $n$ everywhere, i.e., $g$ is an immersion.



          This implies $text{Imm}(M, N)$ is dense in $C^infty(M, N)$. Now for any $f in text{Imm}(M, N)$ consider the map $F : M times M to N times N$, $F(x, y) = (f(x), f(y))$. By a $C^1$-small homotopy we can modify $F$ to $G$ so that $G$ is transverse to the diagonal $Delta_N subset N times N$. Let $G|Delta_M$ be the restriction to the diagonal of $M times M$. As $G$ is $C^1$-close to $F$, image of $G|Delta_M$ fits inside an $epsilon$-neighborhood of $Delta_N subset N times N$, and by $epsilon$-neighborhood theorem we have a normal projection map to $Delta_N$, with which we compose to get a map $g : M to N$. $g times g$ is $C^1$-close to $G$ near $Delta_M$, and since $G pitchfork Delta_N$, by stability of transverse maps, $(g times g) pitchfork Delta_M$ should hold as well. Therefore $g in text{Imm}_pitchfork(M, N)$.



          This implies we have a tower of successively dense subspaces $$text{Imm}_pitchfork(M, N) subset text{Imm}(M, N) subset C^infty(M, N)$$
          In particular, $text{Imm}_pitchfork(M, N)$ is dense in $C^infty(M, N)$, as required. (Note that for the above we need a relative-to-boundary version of this, but once $f$ is already an embedding restricted to (a collar neighborhood of) the boundary, we can do all the required homotopies fixing the boundary.)






          share|cite|improve this answer











          $endgroup$



          Suppose $h : S^1 times I to Sigma$ is a smooth homotopy between the knots $K_0 = h(-, 0)$ and $K_1 = h(-, 1)$. Consider the movie $f : S^1 times I to Sigma times I$ of the homotopy, defined by $f(x, t) = (h(x, t), t)$.



          I claim that $f$ is homotopic, relative to it's boundary, to a map $g$ which is an immersion with transverse double points, i.e., $g$ is an embedding away from finitely many points $x_1, y_1, cdots, x_n, y_n in S^1 times I$ such that $g(x_k) = g(y_k) = p_k$ for $1 leq k leq n$ and $text{im} ,dg_{x_k} pitchfork text{im} , dg_{y_k}$. This follows from a self-transversality theorem that I have stated and given a sketch of proof below.



          As a matter of fact $g$ can be chosen to be $C^1$-close to $f$ as well. $f$ was completely horizontal to $Sigma$, i.e., if we denote $f_t = f|S^1 times {t}$ to be the restriction on the $t$-slice, $f_t' subset (dpi)^*TSigma$ where $pi : Sigma times I to Sigma$ is the projection. Therefore $g_t (=g|S^1 times {t})$ must be $varepsilon$-horizontal to $Sigma$, and in particular, $g_t' notin ker dpi$. This would mean $pi circ g_t$ is an immersion for all $t in I$.



          Note that this is almost enough for what you want: Take $H = pi_Sigma circ g$ to be your new homotopy. This is a homotopy of $K_0$ to $K_1$ through immersions, and the only difficulty is that $H_t$ might stay on a singular knot for an interval's worth of time $t in J subset I$, whereas you want it to stay there for only finitely many points in $I$. Locally near a crossing, $(H_t)_{t in I}$ would look like a pair of undercrossing-overcrossing which gradually grows closer, stays in the position of "$mathsf{X}$" for the interval $t in J$ and then grows apart and becomes a pair of overcrossing-undercrossing. We modify $H$ near those points so that it only stays like an "$mathsf{X}$" for a point $t = t_0 in J$. Then further modify $H$ so that double points are introduced one at a time, i.e, $H_t$ has at most one double point.





          Suppose $M$ is a compact $n$-manifold and $N$ is a compact $2n$-manifold. Let $C^infty(M, N)$ be the space of smooth maps and $text{Imm}(M, N)$ be the subspace of such maps which are immersions.



          Call a map $f in text{Imm}(M, N)$ to be an immersion with clean double points if whenever $x, y in M$, and $f(x) = f(y) = p$, there exists charts $U, V$ around $x, y$ respectively such that $f|_U, f|_V$ are embeddings, $f(U)$ intersects $f(V)$ transversely, and $f(U) cap f(V) = {p}$. Denote the subspace of such maps as $text{Imm}_{pitchfork}(M, N)$



          Theorem (Self-transversality of maps): $text{Imm}_pitchfork(M^n, N^{2n})$ is dense in $C^infty(M^n, N^{2n})$.



          Let $J^1(M, N)$ be the space of $1$-jets of maps $M to N$ and $mathscr{S} subset J^1(M, N)$ be the subset of $1$-jets of non-immersions. The space of $1$-jets is an affine bundle $M_{2n times n}(Bbb R) to J^1(M, N) to M times N$ with fiber keeping track of the formal derivative component. Let $Sigma_{< n} subset M_{2n times n}(Bbb R)$ be the stratified subset of matrices of rank strictly less than $n$. As $mathscr{S}$ fibers over $M times N$ with fiber $Sigma_{< n}$, it is also a stratified subset of $J^1(M, N)$.



          Given any map $f in C^infty(M, N)$, consider it's 1-jet prolongation $j^1 f : M to J^1(M, N)$. By Thom transversality theorem, we can homotope $f$ by a $C^1$-small homotopy to $g in C^infty(M, N)$ such that $j^1 g pitchfork mathscr{S}$. But observe that $dim J^1(M, N) = n(2n+3)$, so $text{codim}, j^1g(M) = dim J^1(M, N)$ $- dim M$ $=$ $2n(n+1)$ whereas $text{codim}, mathscr{S}$ $=$ $text{codim}_{M_{2n times n}(Bbb R)} Sigma_{< n}$ $=$ $n+1$, so $text{codim}, j^1 g(M)$ $+$ $text{codim}, mathscr{S}$ $>$ $dim J^1(M, N)$, forcing $j^1 g$ to be disjoint from $mathscr{S}$. This implies $j^1 g$ has rank $n$ everywhere, i.e., $g$ is an immersion.



          This implies $text{Imm}(M, N)$ is dense in $C^infty(M, N)$. Now for any $f in text{Imm}(M, N)$ consider the map $F : M times M to N times N$, $F(x, y) = (f(x), f(y))$. By a $C^1$-small homotopy we can modify $F$ to $G$ so that $G$ is transverse to the diagonal $Delta_N subset N times N$. Let $G|Delta_M$ be the restriction to the diagonal of $M times M$. As $G$ is $C^1$-close to $F$, image of $G|Delta_M$ fits inside an $epsilon$-neighborhood of $Delta_N subset N times N$, and by $epsilon$-neighborhood theorem we have a normal projection map to $Delta_N$, with which we compose to get a map $g : M to N$. $g times g$ is $C^1$-close to $G$ near $Delta_M$, and since $G pitchfork Delta_N$, by stability of transverse maps, $(g times g) pitchfork Delta_M$ should hold as well. Therefore $g in text{Imm}_pitchfork(M, N)$.



          This implies we have a tower of successively dense subspaces $$text{Imm}_pitchfork(M, N) subset text{Imm}(M, N) subset C^infty(M, N)$$
          In particular, $text{Imm}_pitchfork(M, N)$ is dense in $C^infty(M, N)$, as required. (Note that for the above we need a relative-to-boundary version of this, but once $f$ is already an embedding restricted to (a collar neighborhood of) the boundary, we can do all the required homotopies fixing the boundary.)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 '18 at 17:16

























          answered Dec 6 '18 at 22:21









          Balarka SenBalarka Sen

          10.2k13056




          10.2k13056












          • $begingroup$
            So does this result only hold if $Sigma$ is compact? Or are there analogues of the Thom transversality theorem for maps from compact n-manifolds to general 2n-manifolds?
            $endgroup$
            – Mike Kiss
            Dec 7 '18 at 14:26






          • 1




            $begingroup$
            @MikeKiss Compactness of the codomain is not necessary for transversality results to hold, no. Also note that my argument on the first bit is incomplete; I proposed a possible way to fix it. I'll think about it and see if it works.
            $endgroup$
            – Balarka Sen
            Dec 7 '18 at 15:57


















          • $begingroup$
            So does this result only hold if $Sigma$ is compact? Or are there analogues of the Thom transversality theorem for maps from compact n-manifolds to general 2n-manifolds?
            $endgroup$
            – Mike Kiss
            Dec 7 '18 at 14:26






          • 1




            $begingroup$
            @MikeKiss Compactness of the codomain is not necessary for transversality results to hold, no. Also note that my argument on the first bit is incomplete; I proposed a possible way to fix it. I'll think about it and see if it works.
            $endgroup$
            – Balarka Sen
            Dec 7 '18 at 15:57
















          $begingroup$
          So does this result only hold if $Sigma$ is compact? Or are there analogues of the Thom transversality theorem for maps from compact n-manifolds to general 2n-manifolds?
          $endgroup$
          – Mike Kiss
          Dec 7 '18 at 14:26




          $begingroup$
          So does this result only hold if $Sigma$ is compact? Or are there analogues of the Thom transversality theorem for maps from compact n-manifolds to general 2n-manifolds?
          $endgroup$
          – Mike Kiss
          Dec 7 '18 at 14:26




          1




          1




          $begingroup$
          @MikeKiss Compactness of the codomain is not necessary for transversality results to hold, no. Also note that my argument on the first bit is incomplete; I proposed a possible way to fix it. I'll think about it and see if it works.
          $endgroup$
          – Balarka Sen
          Dec 7 '18 at 15:57




          $begingroup$
          @MikeKiss Compactness of the codomain is not necessary for transversality results to hold, no. Also note that my argument on the first bit is incomplete; I proposed a possible way to fix it. I'll think about it and see if it works.
          $endgroup$
          – Balarka Sen
          Dec 7 '18 at 15:57


















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