A group $G$ is [Hopfian](http://en.wikipedia.org/wiki/Hopfian_group) if every epimorphism $G\to G$ is an isomorphism. A smooth manifold is aspherical if its universal cover is contractible. Are all fundamental groups of aspherical closed smooth manifolds Hopfian?


Perhaps the manifold structure is irrelevant and makes examples harder to construct, so here is another variant that may be more sensible. Let $X$ be a finite CW-complex which is $K(\pi,1)$. If it helps, assume that its top homology is nontrivial. Is $\pi=\pi\_1(X)$ Hopfian?


**Motivation.** Long ago I proved a theorem which is completely useless but sounds very nice: if a manifold $M$ has certain homotopy property, then the Riemannian volume, as a function of a Riemannian metric on $M$, is lower semi-continuous in the Gromov-Hausdorff topology. (And before you laugh at this conclusion, let me mention that it fails for $M=S^3$.)


The required homotopy property is the following: every continuous map $f:M\to M$ which induces an epimorphism of the fundamental group has nonzero (geometric) degree.


This does not sound that nice, and I tried to prove that some known classes of manifolds satisfy it. My best hope was that all *essential* (as in Gromov's "Filling Riemannian manifolds") manifolds do. I could not neither prove nor disprove this and the best approximation was that having a nonzero-degree map $M\to T^n$ or $M\to RP^n$ is sufficient. I never returned to the problem again but it is still interesting to me.


An affirmative answer to the title question would solve the problem for aspherical manifolds. A negative one would not, and in this case the next question may help (although it is probably stupid because I know nothing about the area):


**Question 2.** Let $G$ be a finitely presented group and $f:G\to G$ an epimorphism. It it true that $f$ induces epimorphism in (co)homology (over $\mathbb Z$, $\mathbb Q$ or $\mathbb Z/2$)?



Are fundamental groups of aspherical manifolds Hopfian?
The Baumslag-Solitar group $B(2,3)=\langle a,b\vert ba^2b^{-1}=a^3\rangle$ is not Hopfian. But it has a natural $K(\pi,1)$ given by the double mapping cylinder of $S^1 \rightrightarrows S^1$ where the maps are $z\mapsto z^2$ and $z\mapsto z^3$. This is a finite CW complex.




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Edit: The double mapping cylinder can be constructed like this. Take a circle $S^1$ and a cylinder $S^1\times I$. Glue one end of the cylinder to the circle by the degree 2 map $z\mapsto z^2$, and glue the other end of the cylinder to the circle by the degree 3 map $z\mapsto z^3$. The $a$ in the presentation above is the loop around the circle, while the $b$ is the loop that goes along the cylinder (whose ends have been brought together, forming a loop).


To see that this is a $K(\pi,1)$, you can check that its universal cover is the product $T\_5\times \mathbb{R}$ of the infinite $5$-regular tree $T\_5$ with the line $\mathbb{R}$. (Think about what this CW complex looks like locally: away from the circle where we glued everything together, it's locally a $2$-manifold. At the circle, we have $2+3=5$ half-planes meeting along their edges.) There is a picture of this universal cover on page 3 of Farb-Mosher, "[A rigidity theorem for the solvable Baumslag-Solitar groups](http://www.math.uchicago.edu/~farb/papers.html)", Inventiones 131 2 (1998), 419-451. 


