Question regarding the quotient space. After recently learning about uniformities on topological groups, I've been looking at various problems. I'm having trouble with the following:
If $H$ is a closed subgroup of a topological group $G$, why does it follow that the quotient space $G / H$ is Tychonoff?
 A: Since $H$ is closed, it follows that $G/H$ is Hausdorff and regular, see e.g. Hewitt–Ross, Abstract Harmonic Analysis I, in particular Theorem (5.21), page 38.
It suffices to exhibit a uniform structure on $G/H$ inducing the topology. Let $V \subset G$ be a neighborhood of the identity. Put
$$
U_{V} = \{(xH, yH) \in G/H \times G/H\,:\,yH \subset VxH\}.
$$
The system $\mathfrak{U} = \{U_{V}\,:\,V\subset G\text{ is a neighborhood of the identity}\}$ is a base for a uniform structure on $G/H$:


*

*The diagonal belongs to each set $U_V$.

*We have $yh = vxh'$ if and only if $xh' = v^{-1}yh$, so $(xH,yH) \in U_{V}$ if and only if $(yH,xH) \in U_{V^{-1}}$.

*If $(xH,yH) \in U_{V}$ and $(yH,zH) \in U_{V'}$ then $zH \subset V'yH \subset V'VxH$, so $U_{V'} \circ U_{V} \subset U_{V'V}$. Therefore: If $W$ is a neighborhood of the identity such that $W^2 \subset V$ (such a $W$ exists for every $V$ by continuity of the multiplication) then $U_{W} \circ U_{W} \subset U_{W^2} \subset U_V$.

*Certainly $U_{V \cap V'} \subset U_{V} \cap U_{V'}$.
Finally, the topology induced by the uniform structure has the sets $U_{V}[xH] = \{yH\,:\,yH \subset VxH\}$ as a base, and those describe precisely the usual neighborhood base of the topology of $G/H$. It is not hard to check that the quotient map $\pi:G \to G/H$ is (left) uniformly continuous and that it has the usual quotient property both for continuous and for (left) uniformly continuous maps out of $G$ which are constant on $H$-cosets.
