# Realizing homomorphisms between fundamental groups

Let $X,Y$ be compact connected manifolds and $\varphi\colon\pi_1(X)\to\pi_1(Y)$ be a homomorphism between their fundamental groups. Under what conditions on $X$, $Y$ and $\varphi$ is it true that $\varphi$ is the homomorphism induced by an appropriate continuous map $f\colon X\to Y$?

In general there is an obstruction living in $H^3(X,\pi_2Y)$. Choose a CW structure on $X$ and $Y$ with only one 0-cell. Then you can use $\varphi$ to define a map at the level of 1-skeleta (just by sending every 1-cell $e$ to a cellular representative of $\varphi([e])$). Since $\varphi$ is a map of fundamental groups it respects homotopies between paths so you can extend it to the 2-skeleton (the border every 2-cell has trivial class in $\pi_1X$ so it gets sent to a loop whose class in $\pi_1Y$ is itself trivial). So you have a continous map $f:X^2\to Y$ realizing $\varphi$ as a map of fundamental groups. To extend it you need that the cohomology class of the map sending every 3-cell $e$ of $X$ to $[f(\partial e)]\in\pi_2Y$ is 0 (the addition of a boundary correspond to a modification of $f$ at the 2-skeleton that doesn't change its behaviour on $\pi_1X$). If this obstruction is 0 analogously you can find an obstruction living in $H^4(X,\pi_3Y)$ and so on and so forth. If all groups $H^{n+1}(X,\pi_nY)$ are 0 you can realize your map. A special case is when the universal cover of $Y$ is contractible (i.e. $\pi_iY=0$ for all $i>1$), for example for any hyperbolic manifold.
• It might also be helpful to note that whenever Y has a contractible universal covering space, then any homomorphism $\pi_1(X)\rightarrow\pi_1(Y)$ is induced by a map $X\rightarrow Y$, unique up to homotopy. This is also in Hatcher. – Will Chen May 14 '14 at 16:57
• Thanks for useful answers. Hatcher has it in his book as Proposition 1.b9 if $X$ is a connected CW complex, I found it. Does every compact connected manifold admit a structure of a CW complex? – William of Baskerville May 16 '14 at 13:53
• Every differentiable (or even PL) manifold does. Alternatively you could just replace your manifold with an homotopy equivalent CW complex (from corollary A.12 in Hatcher's book) since your question is not really affected by replacing $X$ with an homotopy equivalent space. – Denis Nardin May 16 '14 at 14:15