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I am looking for some examples of closed, orientable 3-manifolds $M$ and $N$ and a homomorphism $\phi : \pi_1(M) \to \pi_1(N)$ such that $\phi$ is not induced by any continuous map $f : M \to N$. Are there examples of this occurring for lens spaces? I do not believe that this phenomenon can occur for maps between surfaces.

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    $\begingroup$ This cannot happen when $N$ is an Eilenberg-Maclane space of type $K(G,1)$. So yes, it does not happen for a map whose target is a surface, because surfaces are all of type $K(G,1)$ (unless that surface is the sphere or projective plane...). Nor for a map whose target is an irreducible 3-manifold with infinite fundamental group, for the same reason. $\endgroup$ – Lee Mosher Sep 28 '16 at 19:35
  • $\begingroup$ Nice question. I'm wondering what happens when $N$ is the projective 3-space (which is orientable). $\endgroup$ – YCor Oct 1 '16 at 14:01
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    $\begingroup$ ... and would suggest cross-posting to MathOF (please then provide links in both directions) $\endgroup$ – YCor Oct 1 '16 at 22:46
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    $\begingroup$ @YCor Crossposting is discouraged---the correct procedure is to wait for an answer, then flag for migration. $\endgroup$ – Danu Oct 13 '16 at 15:32
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    $\begingroup$ @YCor: The answer is positive when the target is any lens space (this includes the projective space), see my answer below. At the same time, I agree, there is a much better chance for a complete answer at MO. $\endgroup$ – Moishe Kohan Oct 13 '16 at 16:19
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To piggyback on Moishe's answer: Every map $T^3 \to T^3 \# T^3$ has degree zero. Thus one sees immediately by investigating the cohomology that the induced map on $H^1$ is not surjective, and in particular there are many homomorphisms that cannot be realized.

Suppose the map has nonzero degree; make it transverse to the connected sum sphere. Write $T^3 = M_1 \cup M_2$, where $M_i$ is the inverse image of the $i$th component of the connected sum; the boundary $\partial M_1 = \partial M_2 = f^{-1}(S^2)$. Then we have a map $(M_i, \Sigma) \to (T^3 \setminus D^3, S^2)$. Then the map $H_3(M_i, \Sigma) \to H_3(T^3 \setminus D^3, S^2)$ is still multiplication-by-$d$ (degree is an entirely local property!) Collapsing the 2-sphere boundary to get a map $(M_i, \Sigma) \to (T^3, *)$, this is again degree $d$. This implies that $\Sigma$ is not a 2-sphere, since $T^3$ is irreducible, and any map $S^3 \to T^3$ has degree zero.

I claim that this implies $\Sigma$ is incompressible inside the domain. For suppose you had a compressing disc. Because $\pi_2T^3$ is trivial, this implies the compressing disc is homotopic to a disc inside $\Sigma$; but by assumption the original loop was not null in $\Sigma$. This gives a contradiction.

Now an incompressible surface is $\pi_1$-injective by the loop theorem. So it must be a map $T^2 \to T^3$ that's $\pi_1$-injective; one can verify that this means that the surface is homologically nontrivial in $T^3$. But it bounds the $M_i$! So we've arrived at a contradiction to the original claim, and thus every map $T^3 \to T^3 \# T^3$ is degree zero.

This argument is not particularly special to $T^3$; you should be able to mimic it to get many more examples. The general result is that a map $Y \to N$ has degree zero if $N$ has more prime summands than $Y$ has disjoint, non-parallel, incompressible surfaces.

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    $\begingroup$ Right, but there is an easier way to prove that the map has to have zero degree, just note that $\pi_1(T^3)\to \pi_1(T^3\# T^3)$ cannot be onto a finite index subgroup since every finite index subgroup of $G_1 * G_2$ is a nontrivial free product while an abelian group cannot be a nontrivial free product. $\endgroup$ – Moishe Kohan Oct 13 '16 at 18:43
  • $\begingroup$ @Moishe Maybe easier for you, but I am an ignoramus of group theory :) In any case, I think the idea outlined here should provide significantly more examples than a group theoretic argument alone, which should hopefully justify the complication. $\endgroup$ – user98602 Oct 13 '16 at 19:30
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    $\begingroup$ That's fine, it is a very nice answer anyway. Just in case, here is a paper you may want to consider reading during the spare time that you do not have: P.Scott, T.Wall "Topological methods in group theory", math.hunter.cuny.edu/olgak/scott_wall.pdf, where they use methods inspired by low-dimensional topology to prove group-theoretic results. $\endgroup$ – Moishe Kohan Oct 20 '16 at 4:40
  • $\begingroup$ @MoisheCohen Thanks for the link. It looks interesting. $\endgroup$ – user98602 Oct 20 '16 at 4:48
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This is a partial answer to your question, based on this mathoverflow thread. It was observed in the mathoverflow answer that the first obstruction for existence of a continuous map $X\to Y$ (inducing the given homomorphism of fundamental groups) is in $H^3(X, \pi_2(Y))$. In the case when $Y$ is a lens space, $\pi_2(Y)=0$, hence the first obstruction vanishes. If $X$ is 3-dimensional (as in your question), there are no further obstructions. This answers the question about the case of lens spaces.

Thus, in order to find examples when a map between 3-manifolds does not exist you need to consider target manifolds which are nontrivial connected sums (as they have to have $\pi_2\ne 0$). That I am not sure how to do. It seems reasonable to consider maps to $T^3\# T^3$ as a test case.

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