Let $G$ be a topological group. Let $H$ be the closure of the set $\{1\}$. Every continuous map of $G$ to a $T_1$-space factors over the quotient $G/H$
The above is part (b) of Proposition 1.18 in Principles of Harmonic Analysis. Here is the author's proof:
For part (b), let $x \in G$. As the translation by $x$ is a homeomorphism, the closure of the set $\{x\}$ is the set $xH = Hx$. So, if $A \subset G$ is a closed set, then $A = AH=HA$. Let $f : G \to Y$ be a continuous map into a $T_1$-space $Y$. For $y \in Y$, the singleton $\{y\}$ is closed, so $f^{-1}(\{y\})$ is closed, hence of the form $AH$ for some set $A \subset G$. This implies that $f(gh) = f(g)$ for every $g \in G$ and every $h \in H$.
I don't see how the last sentence follows the sentence preceding it.
If $g\in AH=f^{-1}(\{y\})$ and $h\in H$, then $gh\in AH= f^{-1}(\{y\})$ by definition, so $f(gh) =f(g) =y$. Every $g$ is contained in $f^{-1}(\{y\})$ for some $y$ because $f$ is a function (we may take $y=f(g) $). Thus $f(gh) =f(g) $ for all $g\in G$, $h\in H$.