Anyone aware of a nice counterexample to "The fundamental group functor is full?" (Which is...false, right?) and are there a nontrivial subcategories on which its restriction is full?

I.e. Can you think of an example of topological spaces $X$ and $Y$ such that there is a group homomorphism $\phi: \pi_1(X, x_0) \to \pi_1 (Y,y_0)$ such that $\phi$ is not induced by any continuous map $f: X \to Y$ ?

Sometimes every homomorphism is induced by a continuous map: For example every automorphism of $\pi_1(S^1)$ is induced by a map $S^1 \to S^1$. Are all automorphisms of fundamental groups induced by continuous maps? What conditions on spaces $X$ and $Y$ ensure every fundamental group homomorphism is induced by a continuous map?

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    $\begingroup$ It's full on $K(G, 1)$s, I think? $\endgroup$ – Qiaochu Yuan Jun 7 '13 at 18:56
  • $\begingroup$ Is it full on spaces with contractible universal covers? $\endgroup$ – spitespike Jun 11 '13 at 5:05
  • $\begingroup$ I am thinking for a problem. For a basepoint change in pointed space, assuming the space is path connected, we always have base change homomorphism for fundamental groups, can we always have a continuous map in the category of pointed space? If not, I think we can show the functor is not full $\endgroup$ – Minghao Liu Apr 24 '14 at 15:33

Consider $X = \mathbb R P^3$ and $Y = \mathbb R P^2$. They both have $\pi_1$ equal to $\mathbb Z/2\mathbb Z$, but I don't think there is any continuous map $X \to Y$ which induces the identity on $\pi_1$s.

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    $\begingroup$ Indeed, it is easy to see, using the cup product structure, that any continuous map $X\to Y$ induces a trivial map in first cohomology with $\mathbb{Z}_2$ coefficients. The statement in homotopy then follows from the universal coefficient theorems and the naturality of the Hurewicz homomorphism. $\endgroup$ – Miha Habič Jun 7 '13 at 19:22
  • $\begingroup$ @MihaHabič: Dear Miha, Thanks, that's easier than the argument I had in mind. Regards, $\endgroup$ – Matt E Jun 8 '13 at 3:36
  • $\begingroup$ Thanks much. This is excellent as a technical counterexample, but slightly conceptually unsatisfying. Even though the identity from $\pi_1 X \to \pi_1 Y$ is not induced, its inverse is induced by the inclusion map, so in some sense it still arises from a continuous map. Any thoughts on a counterexample without an induced inverse? $\endgroup$ – spitespike Jul 2 '13 at 19:21

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