# How can I complete this proof of a algebraic topology theorem?

The theorem is: Let be f, g : X→Y two continuous maps between topogolical spaces and

H : X$\times$[0, 1] a homotopy such that H(x,0)=f(x) and H(x,1)=g(x).

Given $x_0\in X$ let be $y_0=f(x_0)$, $y_1=g(x_0)$ and $\tau(t)=H(x_0,t)$ a path in Y from $y_0$ and $y_1$.

Then, $g_*=\varphi_{[\tau]}\circ f_*$

i.e. $\require{AMScd}$ \begin{CD} \pi_1(X,x_0) @>f_*>> \pi_1(Y,y_0)\\ @V V V @VV \varphi_{[\tau]} V\\ @>>g_*> \pi_1(Y,y_1) \end{CD}

where

$f_*:\pi_1(X,x_0) \rightarrow \pi_1(Y,y_0),\space f_*([\gamma])=[f\circ \gamma]$,

$g_*:\pi_1(X,x_0) \rightarrow \pi_1(Y,y_1),\space g_*([\gamma])=[g\circ \gamma]$ and

$\varphi_{[\tau]}:\pi_1(Y,y_0) \rightarrow \pi_1(Y,y_1),\space\varphi_{[\tau]}([\gamma])=[\tau^{-1}]·[\gamma]·[\tau]$ In the diagram it's supposed that $\pi_1(X,x_0)$ goes directly to $\pi_1(Y,y_1)$ by $g_*$

$f_*$ and $g_*$ are group homomorphism and $\varphi_{[\tau]}$ is a group isomorphism.

I began supposing

$g_*([\gamma])=\varphi_{[\tau]}\circ f_*([\gamma])=[\tau^{-1}]·[f\circ\gamma]·[\tau] \iff [\tau]·g_*([\gamma])·[\tau^{-1}]·f_*([\gamma]^{-1})=e_{y_0}$

and from here I don't know how to continue, I think there's something with four components which can help me to complete the proof.

Thanks.

## 1 Answer

I'd rather do this explicitly on the level of paths.

A closed path on $(X,x_0)$ is pushed via $f$ to a closed path on $(Y,y_0)$. $\tau$ is a path that starts at $y_0$ and ends at $y_1$. By composing paths $\tau * (f\circ \gamma) * \tau^{-1}$ a closed path $\gamma$ on $(X,x_0)$ is instead pushed to a closed path $(Y,y_1)$.

A homotopy between this path and $g \circ \gamma$ is constructed via $H$. The "pause" function:

$P: [0,1]\times [0,1] \to [0,1]$, $\quad (s,t) \mapsto \begin{cases}s \quad t<s\\t \quad t≥s\end{cases}$

is continuous.

For any s, $(\tau \circ P)(s,t) = \begin{cases}\tau(s) \quad t<s \\ \tau(t) \quad t≥s \end{cases}$ becomes a path that starts at $\tau(s)$ and follows $\tau$ from there until $y_1$. Let $(\tau \circ P)^{-1}(s,-)$ be (for fixed s) the path that runs backwards through $\tau \circ P$.

For fixed $s$ if we put a closed path into $H$, it starts and ends at $\tau(s)$. So for any s the concatenation of $(\tau \circ P)* H(\gamma,s) *(\tau \circ P)^{-1}$ will be a closed path at $y_1$. In the case $s=0$ this path is $\tau * (f\circ \gamma)*\tau^{-1}$, in the case s=1 it is $g \circ \gamma$.

If you show that map is continuous on $I \times I$ you will have constructed a homotopy between these two paths, so they are in the same homotopy class.

If you want to show that the map is continuous explicitly write out the 3 cases in the path concatenation.

A picture of what the homotopy does is that it deforms the $f\circ \gamma$ part of $\tau * (f\circ \gamma) * \tau^{-1}$ into parts of the $g \circ \gamma$ curve via the homotopy between $f$ and $g$ while unzipping the doubled part of path from $\tau$ and $\tau^{-1}$.