Let $G$ and $H$ be two topological groups. Assume that $\phi:\pi\_{1}(G) \to \pi\_{1}(H)$ is a group homomorphism. Is there a continuous function $f:G\to H$ such that $f\_{\*}=\phi$? 



Restriction of "$\pi_{1}$" to topological groups
No. Take $G=SO(5)$ and $H=SO(3)$, both of which have fundamental group $\mathbb{Z}/2$. I claim that there is no continuous map $f: G\to H$ which induces the identity homomorphism.


If there were, then $f$ would induce a nontrivial homomorphism $f\_\ast: H\_1(G;\mathbb{Z}/2)\to H\_1(H;\mathbb{Z}/2)$, and a graded ring homomorphism $f^\ast: H^\ast(H;\mathbb{Z}/2)\to H^\ast(G;\mathbb{Z}/2)$ which is nontrivial on $H^1$. To see that no such ring homomorphism exists, recall (see Hatcher's Algebraic Topology, section 3.D) that $$H^\ast(H;\mathbb{Z}/2)\cong \mathbb{Z}/2[\alpha\_1]/(\alpha\_1^4)$$ and $$H^\ast(G;\mathbb{Z}/2)\cong \mathbb{Z}/2[\beta\_1]/(\beta\_1^8)\otimes\mathbb{Z}/2[\beta\_3]/(\beta\_3^2),$$ where each $\alpha\_i,\beta\_i$ is in degree $i$.


