 Solution

First show that the boundary map is onto and has degree $\pm 1$, and that the fundamental-group injection is actually an isomorphism. An injective map between closed surfaces is a homeomorphism onto an open-and-closed subset by invariance of domain. Thus each component of $\partial M$ maps homeomorphically to a distinct component of $\partial N$.

Orient $M$ and $N$, and writeApply the boundary map in the relative homology sequence. The boundary of the right side has coefficient $d$ on every boundary component of $N$. On the left, an image boundary component has coefficient $+1$ or $-1$, and an omitted component has coefficient zero. Since at least one component is present, $d=\pm 1$; then no component can be omitted and all signs agree. Therefore

Let $H=f_{\ast }{\pi }_1(M)$ and lift $f$ to the connected cover $p:N_H\to N$ corresponding to $H$. If the cover were noncompact, then $H_3(N_H,\partial N_H;\mathbb{Z})=0$. The lifted relative fundamental class would have zero boundary, but its boundary is a nonzero sum of distinct boundary fundamental classes, since the original boundary map is injective and the boundary components are compact. This is impossible. The cover therefore has a finite number $r$ of sheets. Degree multiplicativity givesHence $r=1$. The image of $f_{\ast }$ is the whole target fundamental group, and the assumed injection makes $f_{\ast }$ an isomorphism.