Problem is in the title. I'm not super confident in my approach: could someone let me know if there is anything wrong?

($\Rightarrow$) Let $p: X \rightarrow X$ be the constant path in $A \subset X$ with which $id_X$ is homotopic. $A$ only contains one point $a$ such that $p(x) = a$ for all $x \in X$. We know that $id_X \simeq p$, implying that $id_X \simeq p \circ j$. Similarly, $id_A \simeq p$, implying that $id_A = j \circ p$. Then, we have homotopy equivalence, implying that $X$ has the homotopy type of a one-point space.

($\Leftarrow$) Given that $X$ has the homotopy type of a one-point space, there exists some $f: A \rightarrow X, g: X \rightarrow A$ such that $g \circ f \simeq id_A$ and $f \circ g \simeq id_X$. $f \circ g = p \simeq id_X$. Therefore, we have that $X$ is contractible.


1 Answer 1


Forward direction:

  • I would call $p$ a constant map rather than constant path, since path usually means a map specifically from $[0, 1]$ to a space.
  • What is $j$?
  • The statement $\mathrm{id}_A \simeq p$ doesn't quite make sense, since $p$ is a map $X \to X$ and $\mathrm{id}_A$ is a map $A \to A$. (You could probably also use $p$ to refer to the restriction of $p$ to a map $X \to A$, but this doesn't resolve the issue.)
  • While of course there's nothing mathematically wrong with calling your one-point set $A$, it might contribute to the readability of your proof to simply refer to it as $\{a\}$ (in general, the fewer names/symbols someone has to remember while reading your proof, the easier it is to follow). Even better, make the suggestive choice to call it $\{x\}$ or $\{x_0\}$, since the point lives in $X$ after all.

Reverse direction:

  • You should say what $p$ is (it looks like you want to define it to be $f \circ g$) and why this implies that $X$ is contractible.
  • $\begingroup$ j is the inclusion map — sorry if that was not clear. Thank you for the advice and help! $\endgroup$
    – selfstudy
    Dec 21, 2019 at 18:18

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