There is this last line of a proof that I don't quite understand.
It is taken from Hatcher's Algebraic Topology Chapter 1 page 37.
I understand every line of the proof except the last part.
My question is: why is it when we restrict the homotopy to $t=0$ and $t=1$, we get $\phi_{0*}([f])=\beta_h(\phi_{1*}([f]))$?
The diagram of the proof does not seem to explain in terms of $\phi_{0*}$ and $\beta_h$. This might be an obvious question but I don't get the reason behind that.
Could somebody please give some help on the explanation?
Thanks.
Note that $h_0$ is a loop which is constant at $\varphi_0(x_0)$. Hence setting $t = 0$ in $h_t \cdot (\varphi_t f) \cdot \overline{h_t}$ gives something homotopic to $\varphi_0 f$. For $t = 1$ we have $h_1 = h$ and thus $[h_1 \cdot (\varphi_1 f) \cdot \overline{h_1}] = \beta_h([\varphi_1 f])$ by definition of $\beta_h$.
Therefore, via the homotopy $h_t \cdot (\varphi_t f) \cdot \overline{h_t}$ we have $[\varphi_0 f] = \beta_h([\varphi_1 f])$ or (by definition of $\varphi_{i,*}$) $$\varphi_{0*}([f])=\beta_h(\varphi_{1*}([f])).$$