Conservative functor in a category of co-limit preserving functors between $R$ modules and vector spaces












0












$begingroup$


Let $R$ be a commutative algebra over some field $k$, and let $RM$ denote the category of $R$ modules.



Let $C =$ Fun$_{co}(RM,$ Vect$)$ be the category of colimit preserving functors, where Vect is the category of $k$ vector spaces.



Define a functor $F: C to$ Vect by $F(G) = G(R)$, and so if $eta: G_1 implies G_2$ in $C$, $F(eta) = eta_R$.




Show that $F$ is conservative; that is, $F$ reflects isomorphisms.




My attempt:



First we try to write every $R$ module, $M$, as a cofiltered limit of $R$.



Let $I = {oplus_jR cong N subset M:$ $N$ is a submodule$}$, that is, free submodules of $M$. We equip maps only if they are inclusion maps.



Since $R$ has a copy in $M$, $I$ isn't empty, and given two free submodules we can find another free submodule contained in both. So $I$ is a cofiltered category.



Thus it seems that $M cong$ colim$_IId$.



Next, let $eta: G_1 implies G_2$ be a map in $C$ s.t $eta_R$ is an isomorhism. By definition $G_1, G_2$ preserve colimits and so $G_1(M), G_2(M)$ are cofiltered limits in Vect.



Now I feel pretty close to showing that $eta_M$ is an isomorphism as well, but am not sure how to formally. In a way $G_i(M)$ are made up of $G_i(R)$ in some sense..



Questions:




  1. are my constructions correct?


  2. how can I use $eta_R$ being an isomorphism to show $eta_M$ is as well?











share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Why would $R$ have a "copy" in $M$? And why, given two free submodules, would there be another free submodule containing them?
    $endgroup$
    – Eric Wofsey
    Dec 30 '18 at 2:34
















0












$begingroup$


Let $R$ be a commutative algebra over some field $k$, and let $RM$ denote the category of $R$ modules.



Let $C =$ Fun$_{co}(RM,$ Vect$)$ be the category of colimit preserving functors, where Vect is the category of $k$ vector spaces.



Define a functor $F: C to$ Vect by $F(G) = G(R)$, and so if $eta: G_1 implies G_2$ in $C$, $F(eta) = eta_R$.




Show that $F$ is conservative; that is, $F$ reflects isomorphisms.




My attempt:



First we try to write every $R$ module, $M$, as a cofiltered limit of $R$.



Let $I = {oplus_jR cong N subset M:$ $N$ is a submodule$}$, that is, free submodules of $M$. We equip maps only if they are inclusion maps.



Since $R$ has a copy in $M$, $I$ isn't empty, and given two free submodules we can find another free submodule contained in both. So $I$ is a cofiltered category.



Thus it seems that $M cong$ colim$_IId$.



Next, let $eta: G_1 implies G_2$ be a map in $C$ s.t $eta_R$ is an isomorhism. By definition $G_1, G_2$ preserve colimits and so $G_1(M), G_2(M)$ are cofiltered limits in Vect.



Now I feel pretty close to showing that $eta_M$ is an isomorphism as well, but am not sure how to formally. In a way $G_i(M)$ are made up of $G_i(R)$ in some sense..



Questions:




  1. are my constructions correct?


  2. how can I use $eta_R$ being an isomorphism to show $eta_M$ is as well?











share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Why would $R$ have a "copy" in $M$? And why, given two free submodules, would there be another free submodule containing them?
    $endgroup$
    – Eric Wofsey
    Dec 30 '18 at 2:34














0












0








0





$begingroup$


Let $R$ be a commutative algebra over some field $k$, and let $RM$ denote the category of $R$ modules.



Let $C =$ Fun$_{co}(RM,$ Vect$)$ be the category of colimit preserving functors, where Vect is the category of $k$ vector spaces.



Define a functor $F: C to$ Vect by $F(G) = G(R)$, and so if $eta: G_1 implies G_2$ in $C$, $F(eta) = eta_R$.




Show that $F$ is conservative; that is, $F$ reflects isomorphisms.




My attempt:



First we try to write every $R$ module, $M$, as a cofiltered limit of $R$.



Let $I = {oplus_jR cong N subset M:$ $N$ is a submodule$}$, that is, free submodules of $M$. We equip maps only if they are inclusion maps.



Since $R$ has a copy in $M$, $I$ isn't empty, and given two free submodules we can find another free submodule contained in both. So $I$ is a cofiltered category.



Thus it seems that $M cong$ colim$_IId$.



Next, let $eta: G_1 implies G_2$ be a map in $C$ s.t $eta_R$ is an isomorhism. By definition $G_1, G_2$ preserve colimits and so $G_1(M), G_2(M)$ are cofiltered limits in Vect.



Now I feel pretty close to showing that $eta_M$ is an isomorphism as well, but am not sure how to formally. In a way $G_i(M)$ are made up of $G_i(R)$ in some sense..



Questions:




  1. are my constructions correct?


  2. how can I use $eta_R$ being an isomorphism to show $eta_M$ is as well?











share|cite|improve this question











$endgroup$




Let $R$ be a commutative algebra over some field $k$, and let $RM$ denote the category of $R$ modules.



Let $C =$ Fun$_{co}(RM,$ Vect$)$ be the category of colimit preserving functors, where Vect is the category of $k$ vector spaces.



Define a functor $F: C to$ Vect by $F(G) = G(R)$, and so if $eta: G_1 implies G_2$ in $C$, $F(eta) = eta_R$.




Show that $F$ is conservative; that is, $F$ reflects isomorphisms.




My attempt:



First we try to write every $R$ module, $M$, as a cofiltered limit of $R$.



Let $I = {oplus_jR cong N subset M:$ $N$ is a submodule$}$, that is, free submodules of $M$. We equip maps only if they are inclusion maps.



Since $R$ has a copy in $M$, $I$ isn't empty, and given two free submodules we can find another free submodule contained in both. So $I$ is a cofiltered category.



Thus it seems that $M cong$ colim$_IId$.



Next, let $eta: G_1 implies G_2$ be a map in $C$ s.t $eta_R$ is an isomorhism. By definition $G_1, G_2$ preserve colimits and so $G_1(M), G_2(M)$ are cofiltered limits in Vect.



Now I feel pretty close to showing that $eta_M$ is an isomorphism as well, but am not sure how to formally. In a way $G_i(M)$ are made up of $G_i(R)$ in some sense..



Questions:




  1. are my constructions correct?


  2. how can I use $eta_R$ being an isomorphism to show $eta_M$ is as well?








commutative-algebra category-theory modules limits-colimits






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edited Dec 30 '18 at 1:44







Mariah

















asked Dec 30 '18 at 0:40









MariahMariah

2,1471718




2,1471718








  • 1




    $begingroup$
    Why would $R$ have a "copy" in $M$? And why, given two free submodules, would there be another free submodule containing them?
    $endgroup$
    – Eric Wofsey
    Dec 30 '18 at 2:34














  • 1




    $begingroup$
    Why would $R$ have a "copy" in $M$? And why, given two free submodules, would there be another free submodule containing them?
    $endgroup$
    – Eric Wofsey
    Dec 30 '18 at 2:34








1




1




$begingroup$
Why would $R$ have a "copy" in $M$? And why, given two free submodules, would there be another free submodule containing them?
$endgroup$
– Eric Wofsey
Dec 30 '18 at 2:34




$begingroup$
Why would $R$ have a "copy" in $M$? And why, given two free submodules, would there be another free submodule containing them?
$endgroup$
– Eric Wofsey
Dec 30 '18 at 2:34










1 Answer
1






active

oldest

votes


















3












$begingroup$

Let me start with your second question. There is a very general fact here which is handy.




Theorem: Let $I,C,$ and $D$ be categories. Let $X:Ito C$ be a functor and $c_i:X(i)to A$ be a cocone under $X$. Suppose $G_1,G_2:Cto D$ are functors which both send this cocone to a colimiting cocone, and suppose $eta:G_1to G_2$ is a natural transformation such that $eta_{X(i)}$ is an isomorphism for each $iin I$. Then $eta_A$ is an isomorphism.




Proof: Note that the maps $eta_{X(i)}$ form a natural isomorphism between the functors $G_1circ X$ and $G_2circ X$. They thus induce an isomorphism between the colimits of these functors. But $eta_A$ is exactly the induced morphism between the colimits $G_1(A)$ and $G_2(A)$ (since it commutes with the cocone morphisms by naturality of $eta_A$), so $eta_A$ is an isomorphism. (Explicitly, its inverse can be constructed using the maps $eta_{X(i)}^{-1}$ and the universal property of $G_2(A)$.) $ blacksquare$



So, in your situation where $G_1$ and $G_2$ preserve all colimits, this means that if $eta$ is an isomorphism on every object in some diagram, then $eta$ is also an isomorphism on the colimit of the diagram. In particular, if $eta_R$ is an isomorphism, then $eta_F$ is an isomorphism on any free module $F$, and so $eta$ is an isomorphism on any object which is a colimit of some diagram of free modules.



Now let me turn to your first question. Unfortunately, your work is wrong: most of your assertions are totally unjustified and false in general. Indeed, most modules are not the filtered colimits of their free submodules. For instance, if $Isubset R$ is any nonzero proper ideal then the quotient $M=R/I$ will contain no nonzero free submodules at all.



But, there is no reason to restrict yourself to filtered colimits, or to submodules. Every module is the cokernel of some map between free modules (just consider any presentation of the module). Explicitly, that means any module $M$ can be written as a pushout of a diagram where two objects are free modules and the third object is $0$ (and thus also free). So, as discussed above, this implies $eta_M$ is an isomorphism.



As a final note, I very strongly suspect you have written the definition of the category $C$ incorrectly and you want to consider only the category of $k$-linear colimit preserving functors. It doesn't make a difference for the proof of this particular statement, though.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much! you've helped me a lot in my confusion about this
    $endgroup$
    – Mariah
    Dec 30 '18 at 21:50












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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3












$begingroup$

Let me start with your second question. There is a very general fact here which is handy.




Theorem: Let $I,C,$ and $D$ be categories. Let $X:Ito C$ be a functor and $c_i:X(i)to A$ be a cocone under $X$. Suppose $G_1,G_2:Cto D$ are functors which both send this cocone to a colimiting cocone, and suppose $eta:G_1to G_2$ is a natural transformation such that $eta_{X(i)}$ is an isomorphism for each $iin I$. Then $eta_A$ is an isomorphism.




Proof: Note that the maps $eta_{X(i)}$ form a natural isomorphism between the functors $G_1circ X$ and $G_2circ X$. They thus induce an isomorphism between the colimits of these functors. But $eta_A$ is exactly the induced morphism between the colimits $G_1(A)$ and $G_2(A)$ (since it commutes with the cocone morphisms by naturality of $eta_A$), so $eta_A$ is an isomorphism. (Explicitly, its inverse can be constructed using the maps $eta_{X(i)}^{-1}$ and the universal property of $G_2(A)$.) $ blacksquare$



So, in your situation where $G_1$ and $G_2$ preserve all colimits, this means that if $eta$ is an isomorphism on every object in some diagram, then $eta$ is also an isomorphism on the colimit of the diagram. In particular, if $eta_R$ is an isomorphism, then $eta_F$ is an isomorphism on any free module $F$, and so $eta$ is an isomorphism on any object which is a colimit of some diagram of free modules.



Now let me turn to your first question. Unfortunately, your work is wrong: most of your assertions are totally unjustified and false in general. Indeed, most modules are not the filtered colimits of their free submodules. For instance, if $Isubset R$ is any nonzero proper ideal then the quotient $M=R/I$ will contain no nonzero free submodules at all.



But, there is no reason to restrict yourself to filtered colimits, or to submodules. Every module is the cokernel of some map between free modules (just consider any presentation of the module). Explicitly, that means any module $M$ can be written as a pushout of a diagram where two objects are free modules and the third object is $0$ (and thus also free). So, as discussed above, this implies $eta_M$ is an isomorphism.



As a final note, I very strongly suspect you have written the definition of the category $C$ incorrectly and you want to consider only the category of $k$-linear colimit preserving functors. It doesn't make a difference for the proof of this particular statement, though.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much! you've helped me a lot in my confusion about this
    $endgroup$
    – Mariah
    Dec 30 '18 at 21:50
















3












$begingroup$

Let me start with your second question. There is a very general fact here which is handy.




Theorem: Let $I,C,$ and $D$ be categories. Let $X:Ito C$ be a functor and $c_i:X(i)to A$ be a cocone under $X$. Suppose $G_1,G_2:Cto D$ are functors which both send this cocone to a colimiting cocone, and suppose $eta:G_1to G_2$ is a natural transformation such that $eta_{X(i)}$ is an isomorphism for each $iin I$. Then $eta_A$ is an isomorphism.




Proof: Note that the maps $eta_{X(i)}$ form a natural isomorphism between the functors $G_1circ X$ and $G_2circ X$. They thus induce an isomorphism between the colimits of these functors. But $eta_A$ is exactly the induced morphism between the colimits $G_1(A)$ and $G_2(A)$ (since it commutes with the cocone morphisms by naturality of $eta_A$), so $eta_A$ is an isomorphism. (Explicitly, its inverse can be constructed using the maps $eta_{X(i)}^{-1}$ and the universal property of $G_2(A)$.) $ blacksquare$



So, in your situation where $G_1$ and $G_2$ preserve all colimits, this means that if $eta$ is an isomorphism on every object in some diagram, then $eta$ is also an isomorphism on the colimit of the diagram. In particular, if $eta_R$ is an isomorphism, then $eta_F$ is an isomorphism on any free module $F$, and so $eta$ is an isomorphism on any object which is a colimit of some diagram of free modules.



Now let me turn to your first question. Unfortunately, your work is wrong: most of your assertions are totally unjustified and false in general. Indeed, most modules are not the filtered colimits of their free submodules. For instance, if $Isubset R$ is any nonzero proper ideal then the quotient $M=R/I$ will contain no nonzero free submodules at all.



But, there is no reason to restrict yourself to filtered colimits, or to submodules. Every module is the cokernel of some map between free modules (just consider any presentation of the module). Explicitly, that means any module $M$ can be written as a pushout of a diagram where two objects are free modules and the third object is $0$ (and thus also free). So, as discussed above, this implies $eta_M$ is an isomorphism.



As a final note, I very strongly suspect you have written the definition of the category $C$ incorrectly and you want to consider only the category of $k$-linear colimit preserving functors. It doesn't make a difference for the proof of this particular statement, though.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much! you've helped me a lot in my confusion about this
    $endgroup$
    – Mariah
    Dec 30 '18 at 21:50














3












3








3





$begingroup$

Let me start with your second question. There is a very general fact here which is handy.




Theorem: Let $I,C,$ and $D$ be categories. Let $X:Ito C$ be a functor and $c_i:X(i)to A$ be a cocone under $X$. Suppose $G_1,G_2:Cto D$ are functors which both send this cocone to a colimiting cocone, and suppose $eta:G_1to G_2$ is a natural transformation such that $eta_{X(i)}$ is an isomorphism for each $iin I$. Then $eta_A$ is an isomorphism.




Proof: Note that the maps $eta_{X(i)}$ form a natural isomorphism between the functors $G_1circ X$ and $G_2circ X$. They thus induce an isomorphism between the colimits of these functors. But $eta_A$ is exactly the induced morphism between the colimits $G_1(A)$ and $G_2(A)$ (since it commutes with the cocone morphisms by naturality of $eta_A$), so $eta_A$ is an isomorphism. (Explicitly, its inverse can be constructed using the maps $eta_{X(i)}^{-1}$ and the universal property of $G_2(A)$.) $ blacksquare$



So, in your situation where $G_1$ and $G_2$ preserve all colimits, this means that if $eta$ is an isomorphism on every object in some diagram, then $eta$ is also an isomorphism on the colimit of the diagram. In particular, if $eta_R$ is an isomorphism, then $eta_F$ is an isomorphism on any free module $F$, and so $eta$ is an isomorphism on any object which is a colimit of some diagram of free modules.



Now let me turn to your first question. Unfortunately, your work is wrong: most of your assertions are totally unjustified and false in general. Indeed, most modules are not the filtered colimits of their free submodules. For instance, if $Isubset R$ is any nonzero proper ideal then the quotient $M=R/I$ will contain no nonzero free submodules at all.



But, there is no reason to restrict yourself to filtered colimits, or to submodules. Every module is the cokernel of some map between free modules (just consider any presentation of the module). Explicitly, that means any module $M$ can be written as a pushout of a diagram where two objects are free modules and the third object is $0$ (and thus also free). So, as discussed above, this implies $eta_M$ is an isomorphism.



As a final note, I very strongly suspect you have written the definition of the category $C$ incorrectly and you want to consider only the category of $k$-linear colimit preserving functors. It doesn't make a difference for the proof of this particular statement, though.






share|cite|improve this answer









$endgroup$



Let me start with your second question. There is a very general fact here which is handy.




Theorem: Let $I,C,$ and $D$ be categories. Let $X:Ito C$ be a functor and $c_i:X(i)to A$ be a cocone under $X$. Suppose $G_1,G_2:Cto D$ are functors which both send this cocone to a colimiting cocone, and suppose $eta:G_1to G_2$ is a natural transformation such that $eta_{X(i)}$ is an isomorphism for each $iin I$. Then $eta_A$ is an isomorphism.




Proof: Note that the maps $eta_{X(i)}$ form a natural isomorphism between the functors $G_1circ X$ and $G_2circ X$. They thus induce an isomorphism between the colimits of these functors. But $eta_A$ is exactly the induced morphism between the colimits $G_1(A)$ and $G_2(A)$ (since it commutes with the cocone morphisms by naturality of $eta_A$), so $eta_A$ is an isomorphism. (Explicitly, its inverse can be constructed using the maps $eta_{X(i)}^{-1}$ and the universal property of $G_2(A)$.) $ blacksquare$



So, in your situation where $G_1$ and $G_2$ preserve all colimits, this means that if $eta$ is an isomorphism on every object in some diagram, then $eta$ is also an isomorphism on the colimit of the diagram. In particular, if $eta_R$ is an isomorphism, then $eta_F$ is an isomorphism on any free module $F$, and so $eta$ is an isomorphism on any object which is a colimit of some diagram of free modules.



Now let me turn to your first question. Unfortunately, your work is wrong: most of your assertions are totally unjustified and false in general. Indeed, most modules are not the filtered colimits of their free submodules. For instance, if $Isubset R$ is any nonzero proper ideal then the quotient $M=R/I$ will contain no nonzero free submodules at all.



But, there is no reason to restrict yourself to filtered colimits, or to submodules. Every module is the cokernel of some map between free modules (just consider any presentation of the module). Explicitly, that means any module $M$ can be written as a pushout of a diagram where two objects are free modules and the third object is $0$ (and thus also free). So, as discussed above, this implies $eta_M$ is an isomorphism.



As a final note, I very strongly suspect you have written the definition of the category $C$ incorrectly and you want to consider only the category of $k$-linear colimit preserving functors. It doesn't make a difference for the proof of this particular statement, though.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 30 '18 at 3:27









Eric WofseyEric Wofsey

193k14221352




193k14221352












  • $begingroup$
    thank you very much! you've helped me a lot in my confusion about this
    $endgroup$
    – Mariah
    Dec 30 '18 at 21:50


















  • $begingroup$
    thank you very much! you've helped me a lot in my confusion about this
    $endgroup$
    – Mariah
    Dec 30 '18 at 21:50
















$begingroup$
thank you very much! you've helped me a lot in my confusion about this
$endgroup$
– Mariah
Dec 30 '18 at 21:50




$begingroup$
thank you very much! you've helped me a lot in my confusion about this
$endgroup$
– Mariah
Dec 30 '18 at 21:50


















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