# Homology between a loop and a constant loop [closed]

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?

## closed as off-topic by Greg Martin, Claude Leibovici, Daniel, user91500, user223391 Feb 12 '17 at 3:08

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• What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help. – Greg Martin Feb 9 '17 at 8:58
• @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

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

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

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

$$[f*\overline{g}]=[f*\overline{f}]=[c]$$

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