Two paths $f:I\to X$ and $g:I\to X$ are path homotopic. Does it follow that the loop $f*\bar{g}$ and the constant loop are path homotopic?

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  • @Greg Martin, Let $F$ be a path homology between $f$ and $g$. $f*\bar{f}$ and the constant loop are path homotopic. $f*\bar{f}$ and $f*\bar{g}$ are path homotopic because $F*\bar{f}$ is the path homology between them. It follows that $f*\bar{g}$ and the constant loop are path homotopic. – alch Feb 9 '17 at 9:22
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I assume that $\overline{g}$ is the path defined by $\overline{g}(t)=g(1-t)$ and $*$ is a classical path composition. In that case yes. It follows from the fact the fundamental group $\pi_1(X)$ is well, a group, with following properties:

$$[f][g] = [f*g]$$ $$[f]^{-1} = [\overline{f}]$$

where $[.]$ denotes homotopy class. Now for a constant path $c$ you have


Right, the proof above is for loops only. But the similiar proof can be given in general case.

First of all if $f$ and $g$ are homotopic via $H$ then $\overline{f}$ and $\overline{g}$ are homotopic via $H'(x, t)=H(x, 1-t)$. Now since the homotopy behaves well on composition of paths then


So all you need to know is that if $f_1\sim f_2$ and $g_1\sim g_2$ then $[f_1*g_1]=[f_2*g_2]$ (assuming compatible ends) and $[f*\overline{f}]=[c]$ where $c$ is constant.

  • $[g]$ and $[f]$ do not necessary belong to $\pi_1(X)$ because I didn't say that they are loops. In your proof you assume that $[\bar{f}]$ is equal to $[\bar{g}]$ which is equivalent to my question. – alch Feb 9 '17 at 9:41
  • @alch Oh, my mistake, for some reason I thought these are loops. But I don't assume anywhere that $[\overline{f}]=[\overline{g}]$ (where did you get that from?) even though this is trivial, homotopy given by $H_2(x,t)=H(x, 1-t)$ where $H$ is original homotopy between $f, g$. – freakish Feb 9 '17 at 9:54

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