 Solution

A quotient map $\pi :X\to Y$ is a surjection such that $U\subseteq Y$ is open if and only if ${\pi }^{-1}(U)$ is open in $X$. Thus the topology on $Y$ is exactly the final topology induced by $\pi$.

For an open continuous surjection, continuity gives one implication. For the other, if ${\pi }^{-1}(U)$ is open then $U=\pi ({\pi }^{-1}(U))$ is open by openness and surjectivity. Hence it is a quotient map.

If $\pi :X\to Y$ and $\sigma :Y\to Z$ are quotient maps, they are surjective andThis set is open exactly when ${\sigma }^{-1}(U)$ is open, exactly when $U$ is open. Thus the composition is a quotient map.

For the factorization, continuity of $g$ implies continuity of $f=g\circ \pi$. Conversely, if $f$ is continuous and $U$ is open in $Z$, then ${\pi }^{-1}(g^{-1}(U))=f^{-1}(U)$ is open. The quotient property makes $g^{-1}(U)$ open. This holds for every open $U$, so $g$ is continuous.