Prove that any composition of quotient maps is a quotient map
My Proof
Let $f : X \to Y$ and $g : Y \to Z$ be quotient maps, we show that $g \circ f : X \to Z$ is a quotient map by showing that $Z$ has the quotient topology on it induced by the map $g \circ f$, i.e. that $V \subseteq Z$ is open in $Z$ if and only if $(g \circ f)^{-1}[V]$ is open in $X$
To that end pick $V$ open in $Z$ then $(g \circ f)^{-1}[V]$ is open in $X$ by continuity, completing one direction.
To prove the converse we pick a saturated open set $U'$ in $X$ with respect to $g \circ f$, then $U' = (g \circ f)^{-1}[V']$ for some subset $V' \subseteq Z$, and we need to show $V'$ is open in $Z$. Now $U' = f^{-1}[g^{-1}[V']]$ and put $\Lambda = \ g^{-1}[V']$, then $U' = f^{-1}[\Lambda]$, and since $f$ is a quotient map we have $A$ open in $Y \iff f^{-1}[A]$ is open in $X$, so since $U'$ is open in $X$, we have $\Lambda = \ g^{-1}[V']$ open in $Y$, and since $g$ is a quotient map we have $W$ open in $Z \iff g^{-1}[W]$ is open in $Y$, therefore $V'$ must be open in $Z$. $\square$
Is my proof satisfactory and rigorous enough?
Best Answer
The proof is fine. I would maybe do it like this:
$U \subseteq Z$ open iff $g^{-1}[U] \subseteq Y$ is open as $g$ is quotient.
$g^{-1}[U] \subseteq Y$ is open iff $f^{-1}[g^{-1}[U]] \subseteq X$ open as $f$ is quotient. And as $(g \circ f)^{-1}[U]= f^{-1}[g^{-1}[U]]$ we have
$U \subseteq Z$ open iff $(g \circ f)^{-1}[U]$ open, showing that $g \circ f$ is quotient.