= Solution
{c}
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> and
$$
(\sigma\circ\pi)^{-1}(U)=\pi^{-1}(\sigma^{-1}(U)).
$$
This set is open exactly when $\sigma^{-1}(U)$ is open, exactly when $U$ is open. Thus \b[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 \b[$g$ is continuous].
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