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Let $X$ be a path connected space. Show that $X$ is simply connected iff any two paths in $X$ with the same initial and terminal points are path-homotopic

Recall $X$ simply connected means $\pi_1(X, p) = \{[c_p]\}$, and $X$ is path-connected.

My Proof : Suppose $X$ is simply connected, and let $f, g$ be two paths in $X$ with initial point $p$, and terminal point $q$, then $f \cdot \bar{g} \sim c_p$, let $G$ denote this homotopy, i.e $G: f \cdot \bar{g} \sim c_p$

Define $H(s, t) = G(s, t) \cdot g$, which can be easily verified is a path homotopy between $f$ and $g$

Conversely suppose any two paths in $X$ with the same initial and terminal points, are path-homotopic. Let $h$ be any loop at a point $p \in X$, pick a point $q = h(a)$ for some $a \in [0, 1]$, then $f =h|_{[0, a]}$, and $g =\overline{h|_{[a, 1]}}$ are two paths with the same initial and terminal points, hence $f$ is path homotopic to $g$ and therefore $f \cdot \bar{g}$ is homotopic to $c_p$, and $f \cdot \bar{g} = h$, therefore $h \sim c_p$, and $\pi_1(X)$ is trivial. $\square$

Is this proof satisfactory and rigorous enough? Any comments on my proof writing skills are also greatly appreciated.

Perturbative
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  • There are several questions which are close to yours, did you search? https://math.stackexchange.com/questions/541064/if-all-paths-with-the-same-endpoints-are-homotopic-then-the-space-is-simply-con?rq=1, https://math.stackexchange.com/questions/1472698/are-all-paths-with-the-same-endpoints-homotopic-in-a-simply-connected-region?rq=1 – Lee Mosher Jul 14 '17 at 13:56
  • A simpler proof for the converse would be: given a cycle at $p$, use the assumption on the cycle and the constant path, which are both paths from $p$ to $p$. – Daniel Schepler Aug 03 '17 at 22:57

1 Answers1

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Strictly speaking $G\cdot g$ gives you a path homotopy between $g$ and $f\cdot \overline{g}\cdot g$.

It is clear that you can then collapse $\overline{g}\cdot g$ with a further homotopy, but maybe it is worth adding (!?)

In the converse direction you are implicitly using the same idea. $f\cdot \overline{g}$ homotopies to $f\cdot \overline{f}$ (or to $\overline{g}\cdot g$) and that lets you collapse the latter to the constant path based on $p$.

user463383
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