Trjtdtk

This is a very simple question involving basic definitions.
I want to prove that if $G$ is a topological group, left multiplication $f_acolon gmapsto ag$ 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 $agin G$, 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.

Hint:

For a topological group $G $, the map $f:G×Gto G $ sending $(x,y) $ to $xcdot 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 $m:Gtimes Grightarrow G$, $1:*rightarrow G$, and $(-)^-1:Grightarrow G$ expressing multiplication, identity, and inversion.

Since $Gtimes G$ is the product of topological spaces, it is equipped with continuous projections $pi_G:Gtimes Grightarrow G$ 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.

Hint: if $fcolon Xtimes Yto Z$ is continuous, then it is also coordinatewise continous, i.e. for every $xin X$ and $yin Y$, the functions $f_xcolon Yto Z$, $ymapsto f(x,y)$ and $f_ycolon Xto Z$, $xmapsto f(x,y)$ are both continuous. This follows from the observation that if $Usubseteq Xtimes Y$ 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).