Hello I have the following question. Let $\gamma$ be a path in $X$ then we say that two loops $f_0,f_1$ are homotopic along $\gamma$ iff there is a continuos map $H:I^2 \to X$ s.t $H(t,0) = f_0(t), H(t,1)= f_1(t)$ and if $H(0,s)=H(1,s)= \gamma(s)$
we had then the following claim: $\pi_1(\gamma(f_0)) =f_1 $ if and only if $f_0*\gamma = \gamma *f_1$ (i.e homotopic this is just by direct calculation) if and only if $f_0$,$f_1$ are homotopic along $\gamma$. Why is the last step true?
Edit:
I found the following "proof" in Bredons Topology Proposition 2.4 where he draws this diagram for the proof in the $\rightarrow$ direction
I understand a bit what happens but it seems really unformal to me. I think that we can fill up the blanket part with the homotopy of our assumption, but why do we have given the non blanket part and why is this continuos. And at the upper line why does there $p,f_1,p^{-1}$ have all the same third. $p*f_1*p^{-1}$ should walk in the first half interval $p$ then in the 3/4 interval f_1 and in the last quarter $p^{-1}$ right? And I dont understand the other direction.
I would appreciate very much some help.
A quick note about Bredon's picture: essentially the formalism needed to complete the picture into a proof is nearly identical to what you would use to show that $g*g^{-1}\simeq \text{id}$. In this new situation, $\gamma$ is a path instead of a loop, and the "constant homotopy" that fills the middle triangle in that proof becomes the homotopy along $\gamma$ filling up the trapezoid.
Below I give a sketch of a proof for $(2)\Rightarrow (3)$ that does not involve transforming the given homotopy $f_0*\gamma \simeq \gamma *f_1$ to $\gamma^{-1}*f_0*\gamma = f_1$. The picture it is trying to formalize is as follows:
Note that $\gamma(0)$ is the basepoint of $f_0$ and $\gamma(1)$ is the basepoint of $f_1$. Throughout I will be using nonstandard intervals $I$ for the homotopies to make the formulas slightly nicer, but of course you can transform everything to $[0,1]$. The paths will always have domain $[0,1]$.
We are given a continuous map $H':[-2,2]^2\to X$ such that $$ H'(t,-2)= \begin{cases} f_0(t/2+1) & -2\leq t\leq 0 \\ \gamma(t/2) & 0\leq t\leq 2 \end{cases} \qquad H'(t,2)= \begin{cases} \gamma(t/2+1) & -2\leq t\leq 0 \\ f_1(t/2) & 0\leq t\leq 2 \end{cases} $$
We need a continuous map $H:[-1,1]^2\to X$ such that $H(-1,s)=H(1,s)= \gamma(s)$ and $H(t,-1) = f_0((t+1)/2)$, $H(t,1)= f_1((t+1)/2)$. Here is a construction that works. First, impose the $H(-1,s)$ and $H(1,s)$ condition directly, and for all $-1<t<1$ and $s\in [-1,1]$, define
$$H(t,s) = \begin{cases} H'\left(t+s, \frac{2(s-t)}{2+(t+s)}\right) & t+s<0 \\ H'\left(t+s, \frac{2(s-t)}{2-(t+s)}\right) & t+s\geq 0 \\ \end{cases}$$
To complete the proof we need to check that $H$ is well-defined, continuous, and that $H(t,-1) = f_0((t+1)/2)$, $H(t,1)= f_1((t+1)/2)$
The only real concern for well-definedness is the $y$-coordinate, but note that, e.g. $t+s$ is never $2$ since $t<1$, and if $t+s=2-\varepsilon$ for $\varepsilon>0$ then $|s-t|\leq \varepsilon$ and thus $\left|\frac{2(s-t)}{2-(t+s)}\right|\leq 2$. The other required restrictions on inputs to $H'$ are resolved similarly.
Continuity can be argued using limits, noting in particular that $$\lim_{(t,s)\to (1,s_0)}H(t,s)=H'\left(1+s_0,\frac{-2(1-s_0)}{1-s_0}\right)=\gamma((1+s_0)/2)=H(1,s_0)$$ [This computation is somewhat annoying, since you can't assume that $t+s\geq 0$ the entire time if $s_0=0$, but in that case you can appeal to the gluing lemma.]
Checking the conditions on the top and bottom of the square is routine.
This is essentially a biconditional proof; indeed $(3)\Rightarrow (2)$ is an easier continuity check because we are "squeezing" instead of "unsqueezing" and so the construction of $H'$ from $H$ does not need to separate out $t=\pm 2$.