In "Infinitesimal computations in Topology", Publ IHES, page 318, Dennis Sullivan writes "Recall any selfmapping 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 nonzero Euler characteristic." Why is that true (even the surface case is a mystery to me)?

That is because surfaces of higher genus have a nonzero simplicial volume. It is a general fact that all selfmaps of closed manifolds with nonzero 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 nonzero Euler characteristic have nonzero simplicial volume. Hence, the same argument is likely to apply in many cases which are interesting. 


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 finitesheeted 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 $\beta\in H^{nk}(\tilde{M},\mathbb{Q})$ 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 2handles 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 2handles into $M$, which is possible since each component of $L$ maps to a contractible loop in $M$. Since the cores of the 2handles 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 


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 $\pm 1$. 

