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This is a very simple question involving basic definitions. I want to prove that if $G$ is a topological group, left multiplication $f_a\colon g\mapsto ag$ is a homeomorphism of $G$. Clearly, this map is bijective and it sufficies to show its continuity.

To prove continuity, if $U(ag)$ is an open nhbd of $ag\in G$, there are open nhbds $V(a)$, $W(g)$ s.t. $V(a)W(g)\subseteq U(ag)$, from which I get $aW(g)\subseteq U(ag)$.

Here I stuck. Can I conclude it is open? If yes, why?

Thank you in advance for your help.

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    $\begingroup$ Just notice left multiplication is a restiction of the multiplication function to the subspace $\{a\}\times G$ $\endgroup$ – YuiTo Cheng Mar 25 at 15:28
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    $\begingroup$ In the end, it sufficies to notice that the restriction of a continuous function is continuous. $\endgroup$ – LBJFS Mar 25 at 15:47
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    $\begingroup$ And this is straightforward $\endgroup$ – YuiTo Cheng Mar 25 at 15:50
  • $\begingroup$ Thank you very much! $\endgroup$ – LBJFS Mar 25 at 15:52
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Hint:

For a topological group $G $, the map $f:G×G\to G $ sending $(x,y) $ to $x\cdot y $ is continuous. Now $f_a $ is nothing but $f $ restricted to $\{a\}×G $. So what can you conclude?

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The map $f_a$ is continuous because, by the definition of topological group, the multiplication is continuous.

And the inverse of $f_a$ is $f_{a^{-1}}$, which is continuous by the same reason.

Therefore, $f_a$ is a homeomorphism.

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This is essentially rephrasing the other answer, but in a Category Theoretic perspective:

Let Top be the category of Topological Spaces and continuous maps. A group object in Top is a topological group, ie. a topological space equipped with continuous maps $m:G\times G\rightarrow G$, $1:*\rightarrow G$, and $(-)^{-1}:G\rightarrow G$ expressing multiplication, identity, and inversion.

Since $G\times G$ is the product of topological spaces, it is equipped with continuous projections $\pi_G:G\times G\rightarrow G$ mapping $(g,h)\mapsto g.$

Furthermore, a group G considered as a category is a category where the only object is the group $G$, and morphisms are left (or right - left is what's relevant here) multiplication by elements of $G$. These are isomorphisms since the coset $gG = G$.

Together, we have the desired result.

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  • $\begingroup$ I don't know category theory, but I love different perspectives! Thank you very much. $\endgroup$ – LBJFS Mar 25 at 16:05
  • $\begingroup$ I'm not a category theorist, but I like to think that I'm fairly familiar with the basics. Yet I do not see how the facts you stated give the result. Could you elaborate? $\endgroup$ – tomasz Mar 25 at 16:54
  • $\begingroup$ @tomasz I elaborated a bit. Hopefully this makes more sense. $\endgroup$ – user458276 Mar 25 at 23:11
  • $\begingroup$ I still don't see how it helps. You just added a bunch of defintions (all of which I've known before). The important point here is that the map $G\to G^2$, $g\mapsto (a,g)$ is a morphism. I don't see how it follows. I mean, it would seem reasonable if it did (it would be desirable if for any group object in any concrete category, translation was a morphism), but I don't think it's obvious. $\endgroup$ – tomasz Mar 26 at 11:41
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Hint: if $f\colon X\times Y\to Z$ is continuous, then it is also coordinatewise continous, i.e. for every $x\in X$ and $y\in Y$, the functions $f_x\colon Y\to Z$, $y\mapsto f(x,y)$ and $f_y\colon X\to Z$, $x\mapsto f(x,y)$ are both continuous. This follows from the observation that if $U\subseteq X\times Y$ is open, then so are its cross-sections (which is immediate by the definition of the product topology).

Note that none of the implications mentioned in the preceding paragraph reverse. This is why in general, a semitopological group need not be topological (although there are some very powerful "automatic continuity" theorems for algebraic structures like groups).

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