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?
Thank you  in advance for your help.
 A: 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? 
A: 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.
A: 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.
A: 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).
