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Let $F_0$ and $F_1$ be compact flat manifolds of dimensions $k$ and $m$, respectively, where $k \geq m$. Suppose $f : \pi_1(F_0) \to \pi_1(F_1)$ is a surjective homomorphism. Consider the covering maps $\mathbb{R}^k \overset{\phi_0}{\longrightarrow} \mathbb{T}^k \overset{\phi_1}{\longrightarrow} F_0$ and $\mathbb{R}^m \overset{\psi_0}{\longrightarrow} \mathbb{T}^m \overset{\psi_1}{\longrightarrow} F_1$. Without loss of generality, suppose that $\phi_0$ and $\psi_0$ are the quotient maps by the standard integer lattices $\mathbb{Z}^k$ and $\mathbb{Z}^m$. Denote by $\{ e_1,\ldots,e_k \}$ and $\{ e^1,\ldots,e^m \}$ the standard bases for $\mathbb{R}^k$ and $\mathbb{R}^m$, respectively, and by $\{ [s_1],\ldots,[s_k] \}$ and $\{ [s^1],\ldots,[s^m] \}$ the standard generators for $\pi_1(\mathbb{T}^k)$ and $\pi_1(\mathbb{T}^m)$.

There exist integers $c_{ij}$ such that $f([s_i]) = \sum_j c_{ij} [s^j]$. Let $T : \mathbb{R}^k \to \mathbb{R}^m$ be the linear map defined by $T(e_i) = \sum_j c_{ij} e^j$. It's clear that $T$ descends to a map $\tilde{T} : \mathbb{T}^k \to \mathbb{T}^m$ such that the diagram $\require{AMScd}$ \begin{CD} \mathbb{R}^k @>T>> \mathbb{R}^m\\ @V \phi_0 V V @VV \psi_0 V\\ \mathbb{T}^k @>>\tilde{T}> \mathbb{T}^m \end{CD} commutes. My question is whether there exists a map $\hat{T} : F_0 \to F_1$ such that the diagram \begin{CD} \mathbb{R}^k @>T>> \mathbb{R}^m\\ @V \phi_0 V V @VV \psi_0 V\\ \mathbb{T}^k @>>\tilde{T}> \mathbb{T}^m\\ @V \phi_1 V V @VV \psi_1 V\\ F_0 @>>\hat{T}> F_1 \end{CD} commutes.

What I know: Bieberbach's second theorem states that, if $G$ and $H$ are Bieberbach subgroups of the isometry group $\mathscr{I}(\mathbb{R}^k)$, then for any isomorphism $X : G \to H$ there exists an affine bijection $S : \mathbb{R}^k \to \mathbb{R}^k$ such that $X(\beta) = S \circ \beta \circ S^{-1}$ for all $\beta \in G$. It's not difficult to show that $k = m$ if and only if $f$ is an isomorphism. In this case, it follows that $S = T$ and, therefore, $T$ descends to the desired map $\hat{T}$.

So, more specifically, my question is whether there is a generalization of Bieberbach's second theorem that implies $T$ descends when $m < k$.

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Charles Frohman has given a simple argument for this: Since the universal cover of $F_1$ is contractible, $f$ is the induced homomorphism of a map $F : F_0 \to F_1$. (See also Realizing homomorphisms between fundamental groups.) As suggested in my edit to the original question, the result now follows, as it's a theorem of Eells–Sampson that $F$ is homotopic to a totally geodesic map.

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There is also perhaps a more "hands-on" approach: each of the maps $\phi_1\colon T^k\to F_0$ and $\psi_1\colon T^m\to F_1$ is precisely the projection map corresponding to the quotient by the (free) action of the holonomy group of the base flat manifold on the covering torus (for details, see e.g. Sec 2 in https://arxiv.org/pdf/1705.08431.pdf or Charlap's book, referenced in the former). Thus, a map $T^k \to T^m$ descends to a map $F_0\to F_1$ if and only if it is equivariant with respect to these actions. This is the case in your example, by construction, using the short exact sequences for $\pi_1(F_i)$.

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  • $\begingroup$ At the risk of sounding incompetent: I've spent a good deal of time trying to write an elementary proof of the equivariance of $\tilde{T}$, but I don't see it. $\endgroup$ – James Dibble Nov 14 '18 at 3:53

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