Let $f\colon S_m\rightarrow S_n$ be a continuous map of degree $\pm1$. Then the induced morphism $f_\bullet \colon \pi_1(S_m) \rightarrow \pi_1(S_n)$ is onto.
How can I prove this?
I know that the map $f_\bullet \colon H^1(S_m) \rightarrow H^1(S_n)$ is injective, but I don't see how to conclude.
If it wasn't onto, the map would factor through a covering space of the codomain. If the covering is infinite, the covering space is noncompact hence has $H^2 = 0$, and the degree is zero; if the covering is finite of degree $d$, the covering map induces multiplication-by-$d$ on $H^2$. In either case you cannot possibly achieve degree $\pm 1$.