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 $f\colon X\to Y$?
[Math] Realizing homomorphisms between fundamental groups
at.algebraic-topologygn.general-topology
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No. A continuous homomorphism $S^1\to G$ yields a map $BS^1\to BG$. The space $BS^1$ is homotopy equivalent to $\mathbb CP^\infty$. There is a topological group $G$ such that $BG$ is homotopy equivalent to the sphere $S^2$. A map corresponding to a generator of $\pi_1G=\pi_2BG=H_2S^2$ would give an isomorphism $H^2BG\to H^2BS^1$, but this is incompatible with the cup product.
EDIT: This example is universal in the following sense: A standard way of making a Kan loop group for the suspension of a based simplicial set $K$ is to apply (levelwise) the free group functor from based sets to groups. The realization of this is then the universal example of a topological group $G$ equipped with a continuous map $|K|\to G$. Apply this with $K=S^1$.
EDIT: Yes in the Lie group case. It suffices to consider compact $G$ since a maximal compact subgroup is a deformation retract. Now put a Riemannian structure on $G$ that is left and right invariant, and use that the geodesics are the cosets of the $1$-parameter subgroups, and that in a compact Riemannian manifold every loop is freely homotopic to a closed geodesic.
Suppose you have basepoints $x_0\in X$, $z_0\in Z$ and $p(z_0)=f(x_0)$. The lift $\tilde{f}:X\to Z$ such that $p\circ \tilde{f}=f$ exists and is continuous if and only if
1) $f_{\ast}(\pi_1(X,x_0))\subseteq p_{\ast}(\pi_1(Z,z_0))$ (this is equivalent to $\tilde{f}$ being a well-defined function).
2) For every evenly covered neighborhood $U\subset Y$, there is an open neighborhood $V\subset X$ such that if $\alpha,\beta:([0,1],0)\to (X,x_0)$ are paths with $\alpha(1),\beta(1)\in V$, then the lifts of $f\alpha$ and $f\beta$ starting at $z_0$ end in the same slice of $U$ in $Z$. (this is equivalent to the continuity of $\tilde{f}$)
For arbitrary spaces, this is about as good as it gets. Without more conditions on $X$ to control how paths vary with respect to their endpoints, there is no way to get around having to deal with how the given cover lifts paths. There are some conditions that do provide insight for some non-locally path connected spaces.
Here is a sufficient condition which generalizes local path-connectivity:
Suppose $(PX)_{x_0}$ is the space of paths in $X$ starting at $x_0$ with the compact-open topology and $ev:(PX)_{x_0}\to X$, $ev(\alpha)=\alpha(1)$ is endpoint-evaluation.
Theorem: If $f_{\ast}(\pi_1(X,x_0))\subseteq p_{\ast}(\pi_1(Z,z_0))$ and $ev:(PX)_{x_0}\to X$ is a quotient map, then $\tilde{f}$ exists and is continuous.
For a proof, see Lemma 2.5 and Corollary 2.6 of
J. Brazas, Semicoverings: a generalization of covering space theory, Homology Homotopy Appl. 14 (2012) 33-63.
The proof doesn't require local triviality. To see an example of this generalization in action, consider something like the suspension of a non-discrete, zero-dimensional space (like the Cantor set). Such a space is not locally path connected, but the evaluation map is quotient so lifts are guaranteed to be continuous. The endpoint-evaluation map is not continuous for Zeeman's example that ACL mentions showing that there is not going to be a nice characterization for all spaces.
Best Answer
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.