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For $K(\pi, 1)$'s, the answer is:

Because Euler characteristic is multiplicative under covering spaces.

Edit: As pointed out in the comments, I was assuming the map was $\pi_1$--injective.

Here's an elementary proof of what you want for surfaces:

The fundamental group of $M$ has rank $2g$, where $g$ is the genus of $M$. So the image $H$ of $\pi_1(M)$ has rank at most $2g$.

If the index of $H$ in $\pi_1(M)$ is infinite, then $f$ has degree zero, as it lifts to a map to a noncompact surface.

So we can assume that $H$ has finite index $m$ in $\pi_1(M)$.

By multiplicativity of Euler characteristic under covers, $H$ is the fundamental group of a surface of genus gm - m + 1, which has rank $2gm - 2m +2$. This is a contradiction unless $m = 1$, in which case the map is a homotopy equivalence and the degree is $\pm 1$.

For $K(\pi, 1)$'s, the answer is:

Because Euler characteristic is multiplicative under covering spaces.

Edit: As pointed out in the comments, I was assuming the map was $\pi_1$--injective.

Here's an elementary proof of what you want for surfaces:

The fundamental group of $M$ has rank $2g$, where $g$ is the genus of $M$. So the image $H$ of $\pi_1(M)$ has rank at most $2g$.

If the index of $H$ in $\pi_1(M)$ is infinite, then $f$ has degree zero, as it lifts to a map to a noncompact surface.

So we can assume that $H$ has finite index $m$ in $\pi_1(M)$.

By multiplicativity of Euler characteristic under covers, $H$ is the fundamental group of a surface of genus gm - m + 1, which has rank $2gm - 2m +2$. This is a contradiction unless $m = 1$, in which case the map is surjective on the fundamental group. Since surface groups are hopfian (since they are residually finite), $f$ is injective on the fundamental group, and so $f$ is a homotopy equivalence. So it has degree $\pm 1$.

For $K(\pi, 1)$'s, the answer is:

Because Euler characteristic is multiplicative under covering spaces.

Edit: As pointed out in the comments, I was assuming the map was $\pi_1$--injective.

Here's an elementary argument for surfaces:

The fundamental group of $M$ has rank $2g$, where $g$ is the genus of $M$. So the image $H$ of $\pi_1(M)$ has rank at most $2g$.

If the index of $H$ in $\pi_1(M)$ is infinite, then $f$ has degree zero, as it lifts to a map to a noncompact surface.

So we can assume that $H$ has finite index $m$ in $\pi_1(M)$.

By multiplicativity of Euler characteristic under covers, $H$ is the fundamental group of a surface of genus gm - m + 1, which has rank $2gm - 2m +2$. This is a contradiction unless $m = 1$, in which case the map is a homotopy equivalence and the degree is $\pm 1$.

For $K(\pi, 1)$'s, the answer is:

Because Euler characteristic is multiplicative under covering spaces.

Edit: As pointed out in the comments, I was assuming the map was $\pi_1$--injective.

Here's an elementary proof of what you want for surfaces:

The fundamental group of $M$ has rank $2g$, where $g$ is the genus of $M$. So the image $H$ of $\pi_1(M)$ has rank at most $2g$.

If the index of $H$ in $\pi_1(M)$ is infinite, then $f$ has degree zero, as it lifts to a map to a noncompact surface.

So we can assume that $H$ has finite index $m$ in $\pi_1(M)$.

By multiplicativity of Euler characteristic under covers, $H$ is the fundamental group of a surface of genus gm - m + 1, which has rank $2gm - 2m +2$. This is a contradiction unless $m = 1$, in which case the map is a homotopy equivalence and the degree is $\pm 1$.

For $K(\pi, 1)$'s, the answer is:

Because Euler characteristic is multiplicative under covering spaces.

For $K(\pi, 1)$'s, the answer is:

Because Euler characteristic is multiplicative under covering spaces.

Edit: As pointed out in the comments, I was assuming the map was $\pi_1$--injective.

Here's an elementary argument for surfaces:

The fundamental group of $M$ has rank $2g$, where $g$ is the genus of $M$. So the image $H$ of $\pi_1(M)$ has rank at most $2g$.

If the index of $H$ in $\pi_1(M)$ is infinite, then $f$ has degree zero, as it lifts to a map to a noncompact surface.

So we can assume that $H$ has finite index $m$ in $\pi_1(M)$.

By multiplicativity of Euler characteristic under covers, $H$ is the fundamental group of a surface of genus gm - m + 1, which has rank $2gm - 2m +2$. This is a contradiction unless $m = 1$, in which case the map is a homotopy equivalence and the degree is $\pm 1$.