Argument that if $H$ is normal subgroup of $G$, then $G/H$ is topological group.

Question

Let $G$ be a topological group and $H$ a normal subgroup of $G$. We want to prove that $G/H$ is a topological group. I have found this argument:

We know that canonical projection $\rho\colon G \rightarrow G/H$ is an open map. We also have the following conmutative diagram:

$\require{AMScd}$ \begin{CD} G @>{T_g}>> G\\ @VVV @VVV\\ G/H @>{T_{\rho(g)}}>> G/H \end{CD}

where vertical arrows are for $\rho$ and $T_g$, $T_{\rho(g)}$ are left traslations.

We see at once that translation is continuous on the quotient.

The argument in bold above is the one that I don't understand: Why to show that traslations are continuous implies that the product $$\phi(aH,bH)=aHbH=abH$$ is continuous on $G/H\times G/H$ under product topology?

Answer 1

$\require{AMScd}$ \begin{CD} G \times G @>\cdot_G>> G\\ @V\pi_{G\times G}VV @VV\pi_GV\\ G\times G/H\times H \\ @VfVV @VV\pi_GV\\ G/H\times G/H @>> \cdot_{G/H}> G/H \end{CD}

The map $\cdot_{G/H}$ is continuos if and only if $\cdot_{G/H}\circ f$ is continuos because the function $f$ that maps every $(g_1,g_2)H\times H$ to $(g_1H,g_2H)$ is an homeomorphism (oviously could not be an homeomorphism between topological groups because you don’t know if they are topological groups ).

By property of quotient, $\cdot_{G/H}\circ f$ is continuos if and only if $\cdot_{G/H}\circ f\circ \pi_{G\times G}$ is continuos and you can observe that this map is $\pi_G\circ\cdot_G$ that it is continuos because $\cdot_{G}$ and $\pi_G$ are continuos.

So $\cdot_{G/H}$ is a continuos map. You can use a similar way to prove that $\cdot_{G/H}^{-1}$ is a continuos map and so $G/H$ is a topological group.