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I'm having trouble understanding this question.

We have a path $h$ in $X$ from $x_0$ to $x_1$ and $\bar{h}$ its inverse path. Then a map $\beta_h:\pi_1(X,x_1)\to \pi_1(X,x_0)$ defined by $\beta_h[f]=[h\circ f\circ \bar{h}]$, for every path $f$ in $X$.

The question is to show that $\beta_h$ depends only on the homotopy class of $h$.

Firstly, it says for every path $f$ in $X$, but surely $f$ has to be a loop or you can't form $[h\circ f\circ \bar{h}]$?

And also, I don't understand why it depends on the homotopy class of $h$, when $[h\circ f\circ \bar{h}]$ is the path going from $x_0$ to $x_1$, around $f$, then back to $x_0$, why does the homotopy class of $h$ matter? In general I don't think I fully understand what this map $\beta_h$ is and would like someone to help me out. Thanks

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    They are asking you to show that *only* the homotopy class of $h$ matters. That is, that if $h\sim k$, then $[h\circ f\circ \overline{h}] = [k\circ f \circ\overline{k}]$. So they are not saying the homotopy of $h$ matters, they are saying you get the same result by replacing $h$ with any path that is homotopic to $h$.2012-05-02
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    ahh, that explains my confusion. Thanks!2012-05-02

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