Left multiplication is homeomorphism of topological groups

Question

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.

Answer 1

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.

Answer 2

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?

Answer 3

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.

Answer 4

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).