In what follows, $\tilde{X}$ will denote, as usual notation, the universal covering space of $X$. Let $f:X\to Y$ be a cellular map of CW-complexes such that $f(x)=y$, for vertices $x\in X$, $y\in Y$. Choose $\tilde{x}\in p^{-1}(x)$, $\tilde{y}\in q^{-1}(y)$, where $p:\tilde{X}\to X$ and $q:\tilde{Y}\to Y$ denote the covering maps. Then it is well-known that there exists a unique map $\tilde{f}:\tilde{X}\to \tilde{Y}$ such that $q\tilde{f}=fp$ and $\tilde{f}(\tilde{x})=\tilde{y}$.
My question: Let $g:\tilde{X}\to \tilde{Y}$ be a map such that $g(\tilde{x})=\tilde{y}$, for vertices $x\in X$, $y\in Y$. Under what condition on fundamental groups, there exists a map $f:X\to Y$ such that $qg=fp$ and $f(x)=y$.
The condition is that there exist a homomorphism $\varphi : \pi_1(X) \to \pi_1(Y)$ such that $g(hx) = \varphi(h) g(x)$ where $h \in \pi_1(X)$ acts on the universal cover in the usual way and similarly for $\varphi(h) \in \pi_1(Y)$. Then we quotient by the actions of both fundamental groups and this is precisely the condition necessary for $g$ to respect that quotient and descend to $f$.
A nice example here where this condition can be made very explicit is the case that $X$ and $Y$ are both tori, so $\widetilde{X} = \mathbb{R}^n, \widetilde{Y} = \mathbb{R}^m$, $X = T^n, Y = T^m$, $\pi_1(X) = \mathbb{Z}^n, \pi_1(Y) = \mathbb{Z}^m$, and $x = 0, y = 0$ to keep things simple. Then if $g : \mathbb{R}^n \to \mathbb{R}^m$ is given by multiplication by a matrix in $M_{m \times n}(\mathbb{R})$, the above condition is equivalent to the condition that $g$ actually lies in $M_{m \times n}(\mathbb{Z})$ (so its restriction to $\mathbb{Z}^n$ provides $\varphi$).