Hatcher Lemma 1.19 
Let $f$ be some loop about $x_0$. From what I understand, we want to show that $\varphi_{0*}([f]) = \beta_h\varphi_{1*}([f])$ or $[\varphi_{0} f] = [h \ast (\varphi_{1}f) \ast \overline{h}]$ From what I gather, Hatcher is claiming that $h_t \ast (\varphi_t f) \ast \overline{h}_t$ is a path homotopy between $\varphi_{0} f$ and $h \ast (\varphi_{1}f) \ast \overline{h}$, and I am having trouble seeing this. Specifically, I am having trouble computing $(h_0 \ast (\varphi_0 f) \ast \overline{h}_0)(s)$ and $(h_1 \ast (\varphi_1 f) \ast \overline{h}_1)(s)$, which I am doing to show they are equal to $\varphi_{0}f(s)$ and $(h \ast (\varphi_{1} f) \ast \overline{h})(s)$. 
I think (?) it is clear that each $h_t \ast (\varphi_t f) \ast \overline{h}_t$ is continuous, but it is less certain (for me) that $H(s,t) = (h_t \ast (\varphi_t f) \ast \overline{h}_t)(s)$ is continuous. If one could clarify why both are continuous, that would be appreciated. 
 A: First of all, $h_1*(\varphi_1f)*\overline{h}_1 = h_*(\varphi_1f)*\overline{h}=\beta_h\varphi_1f$, which is what you want. 
Then note that $h_0$ is a constant path, so the other thing you were trying to compute is homotopic to $\varphi_0f$, by some standard lemma about homotopy of paths; so they both are the thing you're trying to compare, it's a good start ! 
Now $H(s,t)$ is  tricky beast, but once you unravel its definition, its continuit becomes clear from the pasting (or gluing or whatever you call it) lemma. Indeed, $H(s,t)$ is defined as (or at least up to a reparametrization of $[0,1]$) : if $s\in [0, \frac{1}{3}]$, it's $h(3ts)$, if $s\in [\frac{1}{3},\frac{2}{3}]$, it's $\varphi_tf (3s-1)$, and if $s\in [\frac{2}{3},1]$, it's $h(t(1-(3s-2)))$. Now on each of these domains, it is clearly continuous : for the first one, it's because $h$ is, and multiplication is continuous; for the second one it's because $f$ is continuous, and $\varphi$ is continuous in two variables, and the third one is the same as the first one.
Now if you look at the intersections there, so $s=\frac{1}{3}$ and $s=\frac{2}{3}$, you see that the different definitions coincide : 
on $\frac{1}{3}$ you have $h(t)$ on the one hand, and $\varphi_tf(0)= \varphi_t(x_0)= h(t)$ (by definition); 
and on $\frac{2}{3}$, you have on the one hand $\varphi_tf(1) = \varphi_t(x_0)=h(t)$, and on the other hand $h(t)$. Therefore the gluing lemma applies. 
(just a sidenote : here it's really apparent that the groupoid formalism makes this kind of study much easier than the group formalism)
