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.

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)