Examples of real functions, satisfying the following conditions, or provide counter-examples
$g circ f$ is injective, but $g$ is not injective.
$g circ f$ is surjective, but $g$ is not surjective.
$g circ f$ is surjective, but $f$ is not surjective.
$f, g$ are not injective, but $g circ f$ is injective.
I am having trouble with 1,2,3 as I am not sure how to prove them.
For 4) I was thinking about doing a contra-positive example (if $g circ f$ is not injective then $f$ and $g$ are injective. This contradicts because we know that injectivity holds through composition but I am not sure if that is the right way to do that question)
Thank you for your help in advance and if you need extra definitions let me know
real-analysis formal-proofs
edited Nov 21 '18 at 20:20
Lorenzo B.
1,8322520
asked Nov 21 '18 at 20:13
smith sam
184
add a comment |
$g circ f$ is injective, but $g$ is not injective.
$g circ f$ is surjective, but $g$ is not surjective.
$g circ f$ is surjective, but $f$ is not surjective.
$f, g$ are not injective, but $g circ f$ is injective.
I am having trouble with 1,2,3 as I am not sure how to prove them.
For 4) I was thinking about doing a contra-positive example (if $g circ f$ is not injective then $f$ and $g$ are injective. This contradicts because we know that injectivity holds through composition but I am not sure if that is the right way to do that question)
Thank you for your help in advance and if you need extra definitions let me know
real-analysis formal-proofs
edited Nov 21 '18 at 20:20
Lorenzo B.
1,8322520
asked Nov 21 '18 at 20:13
smith sam
184
add a comment |
$g circ f$ is injective, but $g$ is not injective.
$g circ f$ is surjective, but $g$ is not surjective.
$g circ f$ is surjective, but $f$ is not surjective.
$f, g$ are not injective, but $g circ f$ is injective.
I am having trouble with 1,2,3 as I am not sure how to prove them.
For 4) I was thinking about doing a contra-positive example (if $g circ f$ is not injective then $f$ and $g$ are injective. This contradicts because we know that injectivity holds through composition but I am not sure if that is the right way to do that question)
Thank you for your help in advance and if you need extra definitions let me know
real-analysis formal-proofs
edited Nov 21 '18 at 20:20
Lorenzo B.
1,8322520
asked Nov 21 '18 at 20:13
smith sam
184
$g circ f$ is injective, but $g$ is not injective.
$g circ f$ is surjective, but $g$ is not surjective.
$g circ f$ is surjective, but $f$ is not surjective.
$f, g$ are not injective, but $g circ f$ is injective.
I am having trouble with 1,2,3 as I am not sure how to prove them.
For 4) I was thinking about doing a contra-positive example (if $g circ f$ is not injective then $f$ and $g$ are injective. This contradicts because we know that injectivity holds through composition but I am not sure if that is the right way to do that question)
Thank you for your help in advance and if you need extra definitions let me know
real-analysis formal-proofs
real-analysis formal-proofs
edited Nov 21 '18 at 20:20
Lorenzo B.
1,8322520
asked Nov 21 '18 at 20:13
smith sam
184
edited Nov 21 '18 at 20:20
Lorenzo B.
1,8322520
asked Nov 21 '18 at 20:13
smith sam
184
edited Nov 21 '18 at 20:20
Lorenzo B.
1,8322520
edited Nov 21 '18 at 20:20
Lorenzo B.
1,8322520
edited Nov 21 '18 at 20:20
Lorenzo B.
1,8322520
1,8322520
asked Nov 21 '18 at 20:13
smith sam
184
asked Nov 21 '18 at 20:13
smith sam
184
asked Nov 21 '18 at 20:13
smith sam
184
184
add a comment |
add a comment |
1 Answer
1
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$$
f:xin{0}rightarrow 0inmathbb R\
g:xinmathbb Rrightarrow 0in mathbb R
$$
then $f$ and $gcirc f:xin{0}rightarrow 0inmathbb R$ are injective because domain is a singleton but $g$ isn't injective.- There aren't such $g$: let $f:Xrightarrow Y, g:Yrightarrow Z$ such that $gcirc f$ is surjective then for every $zin Z$ exists $xin X$ such that $z=(gcirc f)(x)=g[f(x)]$ then let $y=f(x)$ we have $z=g(y)$ and $g$ is necessary surjective.
$$
f:xinmathbb Rrightarrow 0inmathbb R\
g:xinmathbb Rrightarrow 0in {0}
$$
then $g$ and $gcirc f:xinmathbb Rrightarrow 0in{0}$ are surjective because codomain is a singleton but $f$ isn't surjective.
$f$ must be always injective: let $f:Xrightarrow Y, g:Yrightarrow Z$ such that $gcirc f$ is injective then for every $x, x'in X$ such that $f(x)=f(x')$ we have trivially that $g[f(x)]=g[f(x')]$. This implies that $x=x'$ because $gcirc f$ is injective then also $f$ must be injective.
answered Nov 21 '18 at 20:33
P De Donato
4147
add a comment |
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$$
f:xin{0}rightarrow 0inmathbb R\
g:xinmathbb Rrightarrow 0in mathbb R
$$
then $f$ and $gcirc f:xin{0}rightarrow 0inmathbb R$ are injective because domain is a singleton but $g$ isn't injective.- There aren't such $g$: let $f:Xrightarrow Y, g:Yrightarrow Z$ such that $gcirc f$ is surjective then for every $zin Z$ exists $xin X$ such that $z=(gcirc f)(x)=g[f(x)]$ then let $y=f(x)$ we have $z=g(y)$ and $g$ is necessary surjective.
$$
f:xinmathbb Rrightarrow 0inmathbb R\
g:xinmathbb Rrightarrow 0in {0}
$$
then $g$ and $gcirc f:xinmathbb Rrightarrow 0in{0}$ are surjective because codomain is a singleton but $f$ isn't surjective.
$f$ must be always injective: let $f:Xrightarrow Y, g:Yrightarrow Z$ such that $gcirc f$ is injective then for every $x, x'in X$ such that $f(x)=f(x')$ we have trivially that $g[f(x)]=g[f(x')]$. This implies that $x=x'$ because $gcirc f$ is injective then also $f$ must be injective.
answered Nov 21 '18 at 20:33
P De Donato
4147
add a comment |
$$
f:xin{0}rightarrow 0inmathbb R\
g:xinmathbb Rrightarrow 0in mathbb R
$$
then $f$ and $gcirc f:xin{0}rightarrow 0inmathbb R$ are injective because domain is a singleton but $g$ isn't injective.- There aren't such $g$: let $f:Xrightarrow Y, g:Yrightarrow Z$ such that $gcirc f$ is surjective then for every $zin Z$ exists $xin X$ such that $z=(gcirc f)(x)=g[f(x)]$ then let $y=f(x)$ we have $z=g(y)$ and $g$ is necessary surjective.
$$
f:xinmathbb Rrightarrow 0inmathbb R\
g:xinmathbb Rrightarrow 0in {0}
$$
then $g$ and $gcirc f:xinmathbb Rrightarrow 0in{0}$ are surjective because codomain is a singleton but $f$ isn't surjective.
$f$ must be always injective: let $f:Xrightarrow Y, g:Yrightarrow Z$ such that $gcirc f$ is injective then for every $x, x'in X$ such that $f(x)=f(x')$ we have trivially that $g[f(x)]=g[f(x')]$. This implies that $x=x'$ because $gcirc f$ is injective then also $f$ must be injective.
answered Nov 21 '18 at 20:33
P De Donato
4147
add a comment |
$$
f:xin{0}rightarrow 0inmathbb R\
g:xinmathbb Rrightarrow 0in mathbb R
$$
then $f$ and $gcirc f:xin{0}rightarrow 0inmathbb R$ are injective because domain is a singleton but $g$ isn't injective.- There aren't such $g$: let $f:Xrightarrow Y, g:Yrightarrow Z$ such that $gcirc f$ is surjective then for every $zin Z$ exists $xin X$ such that $z=(gcirc f)(x)=g[f(x)]$ then let $y=f(x)$ we have $z=g(y)$ and $g$ is necessary surjective.
$$
f:xinmathbb Rrightarrow 0inmathbb R\
g:xinmathbb Rrightarrow 0in {0}
$$
then $g$ and $gcirc f:xinmathbb Rrightarrow 0in{0}$ are surjective because codomain is a singleton but $f$ isn't surjective.
$f$ must be always injective: let $f:Xrightarrow Y, g:Yrightarrow Z$ such that $gcirc f$ is injective then for every $x, x'in X$ such that $f(x)=f(x')$ we have trivially that $g[f(x)]=g[f(x')]$. This implies that $x=x'$ because $gcirc f$ is injective then also $f$ must be injective.
answered Nov 21 '18 at 20:33
P De Donato
4147
$$
f:xin{0}rightarrow 0inmathbb R\
g:xinmathbb Rrightarrow 0in mathbb R
$$
then $f$ and $gcirc f:xin{0}rightarrow 0inmathbb R$ are injective because domain is a singleton but $g$ isn't injective.- There aren't such $g$: let $f:Xrightarrow Y, g:Yrightarrow Z$ such that $gcirc f$ is surjective then for every $zin Z$ exists $xin X$ such that $z=(gcirc f)(x)=g[f(x)]$ then let $y=f(x)$ we have $z=g(y)$ and $g$ is necessary surjective.
$$
f:xinmathbb Rrightarrow 0inmathbb R\
g:xinmathbb Rrightarrow 0in {0}
$$
then $g$ and $gcirc f:xinmathbb Rrightarrow 0in{0}$ are surjective because codomain is a singleton but $f$ isn't surjective.
$f$ must be always injective: let $f:Xrightarrow Y, g:Yrightarrow Z$ such that $gcirc f$ is injective then for every $x, x'in X$ such that $f(x)=f(x')$ we have trivially that $g[f(x)]=g[f(x')]$. This implies that $x=x'$ because $gcirc f$ is injective then also $f$ must be injective.
answered Nov 21 '18 at 20:33
P De Donato
4147
answered Nov 21 '18 at 20:33
P De Donato
4147
answered Nov 21 '18 at 20:33
P De Donato
4147
answered Nov 21 '18 at 20:33
P De Donato
4147
4147
add a comment |
add a comment |
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