What does Kernel of the natural map mean?

Question

Let $G$ be a topological group and let $H$ be a normal subgroup. Prove that $G/H$ is a topological group, where $G/H$ is regarded as the quotient space of $G$ by the kernel of the natural map.

What does "by the kernel of the natural map" mean?

I understand that the natural map is a map from a space to its quotient space mapping points to their equivalence classes, but I don't understand what they mean by kernel of the natural map since it's talking about a topological space.

I understand "$G/H$ is a quotient space of $G$". This means the natural map $v: G \rightarrow G/H$ through some partition or equivalence relation that results in the space $G/H$.

Answer 1

A priori, the notation $G/H$ could mean two different things in this context:

• (Topological spaces) $G/H$ is the space obtained from $G$ by collapsing all of $H$ to a point (i.e., as a set, quotient of $G$ by the equivalence relation generated by $h \sim h'$ for all $h,h' \in H$);
• (Groups) $G/H$ is the group of left cosets $gH$ (i.e., as a set, quotient of $G$ by the equivalence relation $g \sim g' \iff g^{-1}g' \in H$).

If $G$ is a topological group and $H$ a normal subgroup, then $G/H$ is always to be interpreted in the sense of groups (second bullet point). I think the phrasing of the author meant to indicate that: $G/H$ is an algebraic (group-theoretic) quotient, namely, it is the quotient of $G$ by the kernel $H$ of the canonical projection map $G \to G/H$ that sends an element to its coset.