# Show that X is contractible if and only if X has the homotopy type of one-point space (Munkres 58.5)

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.

• 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.
• 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.