= Solution
First show that the <boundary> map is onto and has degree $\pm1$, 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 write
$$
f_*[M,\partial M]=d[N,\partial N].
$$
Apply the <boundary> map in the relative <homology (mathematics)> 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, $d=\pm1$; then no component can be omitted and all signs agree. Therefore
$$
\boxed{f|_{\partial M}:\partial M\longrightarrow\partial N\text{ is a homeomorphism},\qquad\deg f=\pm1.}
$$
Let $H=f_*\pi_1(M)$ 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 gives
$$
\pm1=\deg f=r\,\deg\widetilde f.
$$
Hence $r=1$. The image of $f_*$ is the whole target <fundamental group>, and the assumed <injection> makes \b[$f_*$ an <isomorphism>].
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