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 $f_{*}=\phi$?
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asked Feb 24, 2014 at 13:07
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2 Answers
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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 $i$.
answered Feb 24, 2014 at 15:43
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$\begingroup$ thank you very much for your interesting answer. what about for further assumption "G and H are path connected abelian topological group"? $\endgroup$ Commented Feb 25, 2014 at 18:21
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$\begingroup$ For path connected abelian Lie groups the answer is yes (because they are all $K(\pi,1)$s). I don't know enough about general topological groups to answer your further question. In fact, I can't think of an example of a path connected abelian topological group which isn't a $K(\pi,1)$. $\endgroup$ Commented Feb 26, 2014 at 6:47
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2$\begingroup$ Topological abelian groups are products of Eilenberg-MacLane spaces so the answer is yes. $\endgroup$– nsrtCommented Feb 26, 2014 at 11:30
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$\begingroup$ @nsrt A group homomorphism between product group is not necessarily a product of group homomorphism. Is not this statement an obstruction for your argument? $\endgroup$ Commented Feb 27, 2014 at 20:23
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1$\begingroup$ If G and H are topological abelian groups, then they are products of Eilenberg-MacLane spaces. So G has a "projection to the $\pi_1$ factor" and H has the "inclusion of the $\pi_1$ factor" i.e. the map you are asking for exists as a composition $G\to K(\pi_1G,1)\stackrel{K(\phi,1)}{\to} K(\pi_1H,1)\to H$. $\endgroup$– nsrtCommented Feb 28, 2014 at 11:15
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Not always. Take $G=H=\mathrm{PGL}(n,\mathbb{R})$, with $n\geq 7$. Every nontrivial continuous homomorphism $G\rightarrow G$ is a Lie automorphism; the group of such automorphisms is generated by inner automorphisms and $A\mapsto {}^tA^{-1} $, so it acts on $\pi _1(G)$ through {$\pm 1$}. On the other hand $\pi _1(G)=\mathbb{Z}/n$ has automorphisms $\neq \pm 1$.
Edit: Sorry, I misread the question - this answers a different question.
