3
$\begingroup$

My definition is that a space $X$ is contractible if it is homotopy equivalent to a point, i.e. there exists $f:X\rightarrow\{pt\}$ and $g:\{pt\}\rightarrow{X}$ such that $f\circ{g}\simeq{id}_{\{pt\}}$ and $g\circ{f}\simeq{id}_X$. I see all over the place (without proof) that a space is contractible if and only if its identity map is nullhomotopic, i.e. there exists a homotopy $F:X\times{I}\rightarrow{X}$ such that $F(x,0)=id_X$ and $F(x,1)$=constant.

I have seen many other statements which would imply this - such as a space $X$ is contractible if and only if every map $f:X\rightarrow{Y}$, for arbitrary $Y$, is nullhomotopic - but all seem to use the statement above as part of the proof. I feel like it should be easy but have got nowhere.

$\endgroup$
4
$\begingroup$

Suppose you have a homotopy $F$ as you describe, and let $x_0\in X$ be the constant you mention (ie $F(x,1)=x_0$ for all $x\in X$).

Take$f:X\to \{x_0\}$ the only possible map and $g: \{x_0\}\to X$ the inclusion map. Then obviously $f\circ g$ is the identity of $\{x_0\}$. Now $g\circ f$ is the constant map $X\to X$, $x\mapsto x_0$, so $F$ is precisely a homotopy between $Id_X$ and $g\circ f$; thus $g\circ f\simeq Id_X$.

Actually you see that the equivalence is just the very definition of what it means that the identity map is homotopic to a constant map, since a constant map is just the composite of a map $X\to \{pt\}$ and a map $\{pt\}\to X$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.