Proving a map in a commutative diagram is continuous.

Question

Suppose $G$ and $H$ are topological groups, with $H \subset G$. I have the following commutative diagram with $f$ being continuous and $p$ being an open surjection (canonical projection). Does this imply that $h$ is continuous?

$\require{AMScd}$ \begin{CD} G @>{f}>> G\\ @VpVV @VVpV\\ G/H @>{h}>> G/H \end{CD}

Take $U$ an open set in $G/H$. I have to show $h^{-1}(U)$ is open in $G/H$, i.e. $p^{-1}(h^{-1}(U))$ is opened in G.

I know $p^{-1}(h^{-1}(U))=(h \circ p)^{-1}(U)=(p \circ f)^{-1}(U)=f^{-1}(p^{-1}(U))$. But does this help to prove the claim?

Answer 1

A map defined on a quotient space $G{/}H$ is continuous iff its composition with the quotient map $p : G \to G{/}H$ is continuous, so $h$ is continuous iff $ h \circ p$ is, and $h \circ p = p \circ f$ by the diagram, and $p$ is continuous by definition, so $h$ is continuous when $f$ is (so we have a composition of continuous maps in $p \circ f$).

Answer 2

You haven't said what the topology is on $G / H$, but the only reasonable topology to use in this situation is the quotient topology, which by definition is the strongest topology such that the projection map $p$ is continuous.

So, since $U \subset G / H$ is open and $p$ is continuous, it follows that $p^{-1}(U) \subset G$ is open.

Next, since $f$ is continuous, it follows that $f^{-1}(p^{-1}(U) \subset G$ is open.

Finally, since from your hypothesis $p$ is an open map, it follows that the set $p(f^{-1}(p^{-1}(U))) = h^{-1}(U) \subset G/H$ is open (that equation between sets uses the hypothesis that $p$ is surjective).