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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 $\overline{h}$ its inverse path. Then a map $\beta _h:\pi_1(X,x_1)\to \pi _1(X,x_0)$ defined by $\beta _h[f]=\left [h\circ f\circ \overline{h}\right ]$, 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 $\left [h\circ f\circ \overline{h}\right ]$?

And also, I don't understand why it depends on the homotopy class of $h$, when $\left [h\circ f\circ \overline{h}\right ]$ 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.

commie trivial
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09867
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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 $h$. – Arturo Magidin May 02 '12 at 16:19
  • ahh, that explains my confusion. Thanks! – 09867 May 02 '12 at 16:28

2 Answers2

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This is effectively an extended comment.

The question is for every path $h$, not $f$. As $[f] \in \pi_1(X, x_1)$, $f$ is a loop in $X$ based at $x_1$, not a path from $x_0$ to $x_1$.

Instead of writing $h\circ f\circ\bar{h}$ you should write $h\cdot f\cdot\bar{h}$ because you are not composing the maps $h$, $f$, and $\bar{h}$, which is what the symbol $\circ$ is usually reserved for. Also, when using composition, we work right to left (i.e. $f\circ g\circ h$ means apply $h$, then apply $g$, then apply $f$), but with concatenation of paths we work left to right (i.e. $f\cdot g\cdot h$ means travel along the path $f$, then the path $g$, then the path $h$).

As Arturo pointed out, you need to show that if $h$ and $k$ are homotopic paths from $x_0$ to $x_1$, then $h\cdot f\cdot\bar{h}$ and $k\cdot f\cdot\bar{k}$ are homotopic loops based at $x_0$.

Michael Albanese
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Actually $\beta_h$ does depend on the homotopy class of $h$. Note that $\beta_h$ is an isomorphism between $\pi_1(X,x_0)$ and $\pi_1(X,x_1)$ but there could be many isomorphisms between the two groups. For example, if $X$ is a Torus and $x_0,x_1$ are two points on it, then there are two ways to move $x_0$ to $x_1$:

enter image description here

The first one corresponds to the identity automorphism of $\mathbb Z\times \mathbb Z$ and the second one corresponds to the automorphism that swaps the coordinates.

Dasheng Wang
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