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I'm wondering about the following:

  1. Every continuous map between smooth manifolds is homotopic to a smooth map.

  2. By density of polynomials in space of continuous functions on [0,1], continuous functions can be approximated by smooth ones with respect to the maximum norm.

Now, assume $G_1$ and $G_2$ are Lie groups; maybe asume $G_1$ and $G_2$ compact. Let $f:G_1\rightarrow G_2$ be a continuous map. Can $f$ approximated (in some sense) by a continuous group homomorphism?

The answer by Dmitri raises another question.

What classes of continuous functions between Lie groups can be approximated by continuous group homomorphisms?

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  • $\begingroup$ It's a bit implicit in answer, but for arbitrary locally compact groups $G_1,G_2$, the set of continuous homomorphisms $\mathrm{Hom}(G_1,G_2)$ is closed in the set of continuous maps (for the topology of uniform convergence on compact subsets). So continuous homomorphisms only approximate continuous homomorphisms (and, if one allows pointwise convergence, they cannot approximate more than group homomorphisms). $\endgroup$
    – YCor
    Commented Jun 10, 2023 at 13:07

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Let $G_1=G_2=U(1)$. Take $f$ such that its image is in an $1/2$-neighborhood of the identity. On the other hand any group homomorphism is either constant or surjective. So there is no chance of approximating continuous maps by homomorphisms.

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  • $\begingroup$ Thanks! could you have a look at my edit? $\endgroup$
    – monoidaltransform
    Commented Jun 8, 2023 at 17:45
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    $\begingroup$ @monoidaltransform you might want to look up for a notion of "quasihomomorphism". there are several definitions, but morally they are all "homomorphisms up to finite distortion" $\endgroup$
    – Dmitrii Korshunov
    Commented Jun 8, 2023 at 17:50
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    $\begingroup$ @monoidaltransform the class of maps that can be approximated arbitrary well by homomorphisms coincides with the class of homomorphisms, since being a homomorphism is a closed condition. $\endgroup$
    – Dmitrii Korshunov
    Commented Jun 8, 2023 at 17:56
  • $\begingroup$ Thanks! Is proving it is a 'closed condition' straightforward? $\endgroup$
    – monoidaltransform
    Commented Jun 8, 2023 at 18:05
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    $\begingroup$ @monoidaltransform yep, $f: G_1\to G_2$ is a homeomorphism iff $f(xy)=f(x)f(y)$ for all $x,y$. For a given pair of points $x,y$ this condition is obviously closed [this is a preimage of identity with respect to the map $f\mapsto f(xy)-f(x)f(y)$]. And an intersection of all such conditions for all pairs $x,y$ is closed too. $\endgroup$
    – Dmitrii Korshunov
    Commented Jun 8, 2023 at 19:09
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Every continuous homomorphism of Lie groups is a real analytic homomorphism. If $G_1$ is connected then $f$ is determined by its derivative at the identity. So there is a finite dimensional family of continuous group homomorphisms, contained in the Lie algebra homomorphisms, a finite dimensional real variety. Not much chance to approximate things. As Dmitry says, you can approximate only homomorphisms.

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