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.


1 Answer 1


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 $\{pt\}\to X$.

  • $\begingroup$ How do you get that $X$ is contractible? (From what you've written) @Captain Lama $\endgroup$
    – Lam18373
    Commented Nov 3, 2022 at 18:39

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