Show that a set of homotopy classes has a single element
$begingroup$
This is from Munkres section 51 problem 2b
Given spaces $X$ and $Y$ let $[X,Y]$ denote the set of homotopy classes of maps of $X$ into $Y$. Show that if $Y$ is path connected, the set $[I,Y]$ has a single element. Here $I=[0,1]$.
My approach to the problem is as follows. Let $f,g in [I,Y]$ and $f,g:I to Y$ be continuous functions. First I define a function $F(x,t)=f(x(1-t))$. Hence $F(x,0)=f(x)$ and $F(x,1)=f(0)$. This show that $F$ is a homotopy from $f$ to $e_{f(0)}$. Likewise, define $G(x,t)=g(xt).$ Here $G(x,1)=g(x)$ and $G(x,0)=g(0)$. This shows that $G$ is a homotopy from $g$ to the constant function $e_{g(0)}$. Since $Y$ is path connected we can find a path that forces a homotopy between $e_{g(0)}$ and $e_{f(0)}$. Hence $fcong e_{f(0)}cong e_{g(0)} cong g$.
Is this the right strategy?
general-topology homotopy-theory
edited Mar 31 '16 at 3:53
Math1000
19k31745
asked Mar 31 '16 at 3:07
emkaemka
2,46973074
$endgroup$
add a comment |
$begingroup$
This is from Munkres section 51 problem 2b
Given spaces $X$ and $Y$ let $[X,Y]$ denote the set of homotopy classes of maps of $X$ into $Y$. Show that if $Y$ is path connected, the set $[I,Y]$ has a single element. Here $I=[0,1]$.
My approach to the problem is as follows. Let $f,g in [I,Y]$ and $f,g:I to Y$ be continuous functions. First I define a function $F(x,t)=f(x(1-t))$. Hence $F(x,0)=f(x)$ and $F(x,1)=f(0)$. This show that $F$ is a homotopy from $f$ to $e_{f(0)}$. Likewise, define $G(x,t)=g(xt).$ Here $G(x,1)=g(x)$ and $G(x,0)=g(0)$. This shows that $G$ is a homotopy from $g$ to the constant function $e_{g(0)}$. Since $Y$ is path connected we can find a path that forces a homotopy between $e_{g(0)}$ and $e_{f(0)}$. Hence $fcong e_{f(0)}cong e_{g(0)} cong g$.
Is this the right strategy?
general-topology homotopy-theory
edited Mar 31 '16 at 3:53
Math1000
19k31745
asked Mar 31 '16 at 3:07
emkaemka
2,46973074
$endgroup$
2
$begingroup$
Yeah. This is a good strategy. To really sell it, wrap it all together in one homotopy.
$endgroup$
– Alex S
Mar 31 '16 at 3:50
2
$begingroup$
To be precise, $[X,Y]$ is the set of homotopy classes of continuous maps $Xto Y$.
$endgroup$
– Math1000
Mar 31 '16 at 3:54
add a comment |
$begingroup$
This is from Munkres section 51 problem 2b
Given spaces $X$ and $Y$ let $[X,Y]$ denote the set of homotopy classes of maps of $X$ into $Y$. Show that if $Y$ is path connected, the set $[I,Y]$ has a single element. Here $I=[0,1]$.
My approach to the problem is as follows. Let $f,g in [I,Y]$ and $f,g:I to Y$ be continuous functions. First I define a function $F(x,t)=f(x(1-t))$. Hence $F(x,0)=f(x)$ and $F(x,1)=f(0)$. This show that $F$ is a homotopy from $f$ to $e_{f(0)}$. Likewise, define $G(x,t)=g(xt).$ Here $G(x,1)=g(x)$ and $G(x,0)=g(0)$. This shows that $G$ is a homotopy from $g$ to the constant function $e_{g(0)}$. Since $Y$ is path connected we can find a path that forces a homotopy between $e_{g(0)}$ and $e_{f(0)}$. Hence $fcong e_{f(0)}cong e_{g(0)} cong g$.
Is this the right strategy?
general-topology homotopy-theory
edited Mar 31 '16 at 3:53
Math1000
19k31745
asked Mar 31 '16 at 3:07
emkaemka
2,46973074
$endgroup$
This is from Munkres section 51 problem 2b
Given spaces $X$ and $Y$ let $[X,Y]$ denote the set of homotopy classes of maps of $X$ into $Y$. Show that if $Y$ is path connected, the set $[I,Y]$ has a single element. Here $I=[0,1]$.
My approach to the problem is as follows. Let $f,g in [I,Y]$ and $f,g:I to Y$ be continuous functions. First I define a function $F(x,t)=f(x(1-t))$. Hence $F(x,0)=f(x)$ and $F(x,1)=f(0)$. This show that $F$ is a homotopy from $f$ to $e_{f(0)}$. Likewise, define $G(x,t)=g(xt).$ Here $G(x,1)=g(x)$ and $G(x,0)=g(0)$. This shows that $G$ is a homotopy from $g$ to the constant function $e_{g(0)}$. Since $Y$ is path connected we can find a path that forces a homotopy between $e_{g(0)}$ and $e_{f(0)}$. Hence $fcong e_{f(0)}cong e_{g(0)} cong g$.
Is this the right strategy?
general-topology homotopy-theory
general-topology homotopy-theory
edited Mar 31 '16 at 3:53
Math1000
19k31745
asked Mar 31 '16 at 3:07
emkaemka
2,46973074
edited Mar 31 '16 at 3:53
Math1000
19k31745
asked Mar 31 '16 at 3:07
emkaemka
2,46973074
edited Mar 31 '16 at 3:53
Math1000
19k31745
edited Mar 31 '16 at 3:53
Math1000
19k31745
edited Mar 31 '16 at 3:53
Math1000
19k31745
19k31745
asked Mar 31 '16 at 3:07
emkaemka
2,46973074
asked Mar 31 '16 at 3:07
emkaemka
2,46973074
asked Mar 31 '16 at 3:07
emkaemka
2,46973074
2,46973074
2
$begingroup$
Yeah. This is a good strategy. To really sell it, wrap it all together in one homotopy.
$endgroup$
– Alex S
Mar 31 '16 at 3:50
2
$begingroup$
To be precise, $[X,Y]$ is the set of homotopy classes of continuous maps $Xto Y$.
$endgroup$
– Math1000
Mar 31 '16 at 3:54
add a comment |
2
$begingroup$
Yeah. This is a good strategy. To really sell it, wrap it all together in one homotopy.
$endgroup$
– Alex S
Mar 31 '16 at 3:50
2
$begingroup$
To be precise, $[X,Y]$ is the set of homotopy classes of continuous maps $Xto Y$.
$endgroup$
– Math1000
Mar 31 '16 at 3:54
2
2
$begingroup$
Yeah. This is a good strategy. To really sell it, wrap it all together in one homotopy.
$endgroup$
– Alex S
Mar 31 '16 at 3:50
$begingroup$
Yeah. This is a good strategy. To really sell it, wrap it all together in one homotopy.
$endgroup$
– Alex S
Mar 31 '16 at 3:50
2
2
$begingroup$
To be precise, $[X,Y]$ is the set of homotopy classes of continuous maps $Xto Y$.
$endgroup$
– Math1000
Mar 31 '16 at 3:54
$begingroup$
To be precise, $[X,Y]$ is the set of homotopy classes of continuous maps $Xto Y$.
$endgroup$
– Math1000
Mar 31 '16 at 3:54
add a comment |
1 Answer
1
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$begingroup$
Your proof is correct!
Just one comment, though: perhaps you can mention why the maps $F$ and $G$ are continuous, for the sake of more clarity.
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Your proof is correct!
Just one comment, though: perhaps you can mention why the maps $F$ and $G$ are continuous, for the sake of more clarity.
$endgroup$
add a comment |
$begingroup$
Your proof is correct!
Just one comment, though: perhaps you can mention why the maps $F$ and $G$ are continuous, for the sake of more clarity.
$endgroup$
add a comment |
$begingroup$
Your proof is correct!
Just one comment, though: perhaps you can mention why the maps $F$ and $G$ are continuous, for the sake of more clarity.
$endgroup$
Your proof is correct!
Just one comment, though: perhaps you can mention why the maps $F$ and $G$ are continuous, for the sake of more clarity.
answered Dec 4 '18 at 7:49
community wiki
Brahadeesh
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2
$begingroup$
Yeah. This is a good strategy. To really sell it, wrap it all together in one homotopy.
$endgroup$
– Alex S
Mar 31 '16 at 3:50
2
$begingroup$
To be precise, $[X,Y]$ is the set of homotopy classes of continuous maps $Xto Y$.
$endgroup$
– Math1000
Mar 31 '16 at 3:54