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 $f\colon X\to Y$?



Realizing homomorphisms between fundamental groups
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.


