Continuous function and induced homomorphism We know that given any continuous map $f : X \to Y$, there exists an induced homomorphism $f_* : \pi_1(X, x) \to \pi_1(Y, f(x))$.
Is the converse true? That is, given any homomorphism $\phi: \pi_1(X, x) \to \pi_1(Y, y)$, is there a continuous map $f : X \to Y$ such that $\phi = f_*$ and $f(x) = y$?
If not, are there conditions for $X$ and $Y$ for the converse to be true?
 A: Here is a really simple example.
Take the real projective plane $\mathbb{R}P^2$ and attach a 3-cell to obtain $\mathbb{R}P^3$. For cellular reasons the inclusion $\mathbb{R}P^2\hookrightarrow\mathbb{R}P^3$ induces an isomorphism $\pi_1\mathbb{R}P^2\xrightarrow\cong\pi_1\mathbb{R}P^3$, both groups being isomorphic to $\mathbb{Z}_2$. However this isomorphism cannot be induced by a map $f:\mathbb{R}P^3\rightarrow\mathbb{R}P^2$ which points in the opposite direction.
For suppose that $f$ exists and is non-trivial on $\pi_1$. Then $f$ induces an isomorphism on $H_1$ by the Hurewitz Theorem, and so also an isomorphism on $H^1(-;\mathbb{Z}_2)$ by the Universal Coefficient Theorem. Now, if $x\in H^1(\mathbb{R}P^3;\mathbb{Z}_2)\cong\mathbb{Z}_2$ is a generator, then $x^3$ generates $H^3(\mathbb{R}P^3;\mathbb{Z}_2)\cong\mathbb{Z}_2$. But if $y\in H^1(\mathbb{R}P^2;\mathbb{Z}_2)$ is a generator, then $f^*y=x$ by the above, giving $x^3=(f^*y)^3=f^*(y^3)=0$ for dimensional reasons. This is a contradiction.
