Degrees of self-maps of aspherical manifolds In "Infinitesimal computations in Topology", Publ IHES, page 318, Dennis Sullivan writes "Recall any self-mapping 
of a Riemann surface of genus $g>1$ either has
degree $0$ or degree $\pm 1$." There is a footnote to that sentence, saying this statement is more generally 
"[t]rue for a closed $K(\pi,1)$-manifold of non-zero Euler characteristic." Why is that true (even the surface case is a mystery to me)?
 A: Here's an argument that the map on fundamental groups $\phi:M\to M$ is surjective if $deg(\phi)\neq 0$, where $M$ is an $n$-dimensional closed  orientable manifold and $\chi(M)\neq 0$. Suppose $deg(\phi)\neq 0$, then there exists a finite-sheeted cover $\tilde{M}\to M$ such that $\phi_{\#}(\pi_1(M))=\pi_1(\tilde{M})$ (as noted by Richard Kent, if $\tilde{M}\to M$ were an infinite cover, then $deg(\phi)=0$). Consider the lift $\tilde{\phi}:M\to\tilde{M}$. Then $deg(\tilde{\phi})\neq 0$ as well. Then $\tilde{\phi}^\ast:H^n(\tilde{M},\mathbb{Q})\to H^n(M,\mathbb{Q})$ is an isomorphism of vector spaces. By Poincare duality, for any $\alpha\in H^k(\tilde{M},\mathbb{Q})$ there exists $\beta\in H^{n-k}(\tilde{M},\mathbb{Q})$ such that $\alpha\cup\beta = [\tilde{M}]$. Then $\tilde{\phi}^\ast(\alpha\cup\beta)=\tilde{\phi}^\ast(\alpha)\cup\tilde{\phi}^{\ast}(\beta) = \tilde{\phi}^\ast[\tilde{M}]\neq 0$, so $\tilde{\phi}^\ast(\alpha)\neq 0$. Thus $\tilde{\phi}^\ast$ is an injection from $H^\ast(\tilde{M},\mathbb{Q})\hookrightarrow H^\ast(M,\mathbb{Q})$. But the covering projection $\tilde{M}\to M$ induces an injection $H^\ast(M,\mathbb{Q})\hookrightarrow H^\ast(\tilde{M},\mathbb{Q})$, so we see that $H^\ast(\tilde{M},\mathbb{Q})\cong H^\ast(M,\mathbb{Q})$ (as graded vector spaces), and therefore $\chi(\tilde{M})=\chi(M)$. Thus the cover $\tilde{M}\to M$ is degree one, and we see that $\phi_{\#}:\pi_1(M)\to \pi_1(M)$ is a surjection.  
If $\pi_1(M)$ is Hopfian, then $\phi_{\#}$ is an isomorphism, and we conclude that $\phi$ is a homotopy equivalence when $M$ is a $K(\pi,1)$, and therefore $deg(\phi)=\pm 1$. However, $\pi_1(M)$ might not be Hopfian. Assume $n\geq 4$ (since $n=2$ is Hopfian, and is taken care of in Richard Kent's answer) and $M$ is aspherical. Then $Ker(\phi_{\#})$ is finitely normally generated (since $\pi_1(M)$ is finitely presented). Choose a link $L\subset M$ such that $Ker(\phi_{\#})$ is normally generated by $\pi_1$ of the components of $L$. We surger $M$ by adding 2-handles along the components of $L$ to get $M'$ such that $\pi_1(M')=\pi_1(M)$ ($M'$ might not be aspherical). But we may extend the map $\phi_{|M-\mathcal{N}(L)}:M-\mathcal{N}(L)\to M$ to a map $\phi': M' \to M$ by mapping the attached 2-handles into $M$, which is possible since each component of $L$ maps to a contractible loop in $M$. Since the cores of the 2-handles are codimension $\geq 2$, we see that $deg(\phi')=deg(\phi)$. But since $\phi'_{\#}:\pi_1(M')\to \pi_1(M)$ is an isomorphism, $\phi$ is homotopic to the classifying map, there's a gap here so we conclude that $deg(\phi')=\pm 1$, so $deg(\phi)=\pm 1$.  
A: 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 $\pm 1$.
A: That is because surfaces of higher genus have a non-zero simplicial volume. It is a general fact that all selfmaps of closed manifolds with non-zero simplicial volume have degree either $-1,0$ or $+1$. (see for example here)
I do not know about the other cases which are mentioned in the footnote.
EDIT: As Bruno Martelli pointed out in a comment, Gromov conjectured that closed aspherical manifolds of non-zero Euler characteristic have non-zero simplicial volume. Hence, the same argument is likely to apply in many cases which are interesting.
