We consider a locally compact abelian group $G$. We equip the real vector space $A(G)$ of continuous group homomorphisms $\mathbb{R}\to G$ with the topology of uniform convergence on compact subsets of $\mathbb{R}$. Then we can define the exponential map $\exp\colon A(G)\to G,u\mapsto u(1)$. It is clear that the image of $\exp$ is contained in the path-component $G_0$ of the identity of $G$. Is it true that the map $\exp\colon A(G)\to G_0$ is a local homeomorphism?
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No. Suppose that $G=(\mathbf{R}/\mathbf{Z})^\mathbf{N}$. Then $\mathrm{Hom}(\mathbf{R},G)=\mathbf{R}^\mathbf{N}$ is not locally compact, hence cannot be locally homeomorphic to $G_0=G$.
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$\begingroup$ Ah, yes! Thanks! Do you know if we can at least say that it is locally injective/bijective/surjective (which seems to be the case in your example)? $\endgroup$– HansCommented May 31 at 11:45
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2$\begingroup$ No, it's not locally injective in my example. Let $f_n$ be the map mapping $x$ to the sequence whose $n$-coordinate is $x\pmod 1$ and others are zero. Then $f_n\neq 0$ tends to $0$ and $\exp(f_n)=0$. In other words, the kernel $\mathbf{Z}^\mathbf{N}$ is not discrete in $\mathbf{R}^\mathbf{N}$. I'm not sure what you mean by "locally surjective". Do you just mean an open map? $\endgroup$– YCorCommented May 31 at 12:33
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$\begingroup$ Thanks! By locally surjective I mean that the image of every nonempty open set has nonempty interior. Not sure whether this implies the map being open. $\endgroup$– HansCommented May 31 at 12:36
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$\begingroup$ For a continuous homomorphism of topological groups, this is equivalent. For an arbitrary continuous map, this is weaker (e.g. $(x,y)\mapsto (x,xy)$ in the plane) — but I don't think it can be called "locally surjective". $\endgroup$– YCorCommented May 31 at 12:48
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1$\begingroup$ @Hans It seems you are caught up in thinking infinite-dimensional groups might behave like Lie groups if they are compact. This is not so. It isn't the case that people have formulated the theorems about Lie groups "conservatively', it really is the case that a general compact group doesn't behave at all like a Lie group, even if connected. Once you leave the compact setting, the behaviour of the exponential map is even wilder, e.g. in the case of $\mathrm{Diff}(S^1)$ or the unitary group of an infinite-dimensional Hilbert space in the strong/weak operator topology. $\endgroup$ Commented May 31 at 23:10