# Left multiplication is homeomorphism of topological groups

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?

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

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?

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.

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.

• I don't know category theory, but I love different perspectives! Thank you very much. – LBJFS Mar 25 at 16:05
• 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? – tomasz Mar 25 at 16:54
• @tomasz I elaborated a bit. Hopefully this makes more sense. – user458276 Mar 25 at 23:11
• 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. – tomasz Mar 26 at 11:41

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).