Proof:

Since $\pi_1(S^1, 1)$ is abelian, the abelianization $\Pi:\pi_1(S^1, 1) \longrightarrow H_1(S^1)$ is an isomorphism and hence maps a generator of $\pi_1(S^1, 1)$ to a generator of $H_1(S^1)$. Since (36.2) represents a generator for $\pi_1(S^1, 1)$ we infer that the cycle (36.2) is a generator for $H_1(S^1)$. We deduce from the diagram (32.1) that

$$f_* = \Pi^{-1}\circ H_1(f)\circ \Pi. \eqno(36.3)$$

From (12.6) and (36.3) we see that

$$H_1(f)\Pi[\eta] = (\Pi\circ f_*)[\eta] = \Pi(($$deg$$\;f)[\eta]) = ($$deg$$\;f)\Pi[\eta].$$

Appealing to the definition (36.1) we see that $m =$   deg$\;f$.