An odd map having odd degree little issue with the proof

Question

In the proof below of hatcher I agree that $Im(\tau) \subset Ker(p_{\#})$. However, what I don't understand why is the other inclusion true ?

Answer 1

$\tau$, here, is precisely the map that takes a singular simplex $\sigma \in C_n(X;\Bbb Z/2)$ and sends it to both its covers in $C_n(\tilde X;\Bbb Z/2)$; that is, there are precisely two maps $\tilde \sigma_i: \Delta^n \to \tilde X$, and we send $\tau(\sigma) = \tilde \sigma_1 + \tilde \sigma_2$.

Now let $\sigma = \sum \sigma_n \in C_n(\tilde X;\Bbb Z/2)$, where the $\sigma_i$ are singular simplices (which we may assume are not repeated) and assume $p_\# \sigma = 0$, which is to say that $\sum a_n p_\# \sigma_n = 0$.

Note that if $\eta_1, \eta_2$ are maps $\Delta^n \to \tilde X$, then $p_\# \eta_1 = p_\# \eta_2$ if and only if either $\eta_1 = \eta_2$ or $\eta_1 = i_* \eta_2$, where $i_*$ is the nontrivial deck transformation $\tilde X \to \tilde X$. So if $\sum p_\# \sigma_n = 0$, then each $p_\# \sigma_n$ pairs off with another one; because we assume none are repated, this means our sum was actually of the form $\sum (\sigma_j + i^*\sigma_j)$, and hence was the image of $\tau(\sum p_\#\sigma_j)$, as desired.