I am reading the book 'Algebraic Topology' by T.T. Dieck, and am confused by one of his results in the opening chapter. He claims:
Proposition 1.2.3
Let $f:X\to Y$ be a surjective continuous map between spaces. Then the following are equivalent:
1) $f$ is a quotient map
2) A set map (i.e. not necessarily continuous) $g:Y\to Z$ to any other topological space $Z$ is continuous if and only if $gf$ is continuous.
So $1)\Rightarrow 2)$ is easy. For $2)\Rightarrow 1)$, if you set $Z=Y$ and take $g=Id_Y$, then $gf=f$ is continuous since $Id_Y$ is continuous. It therefore remains to show $U$ is open in $Y$ if the preimage $f^{-1}(U)$ is open in $X$. However, here I am stuck. Obviously if $f$ is an open map then the result is immediate, but I don't think Dieck is assuming this.
Applying (2) for a $g$ which is known to be continuous will not suffice.
However, we can correct your reasoning with $g=id_Y$ by considering once the topology given on $Y$, and secondly the quotient topology induced by $f$, put on $Z$ (where $Z=Y$ as set).
Since $f:X\to Y$ is continuous and surjective, $Y$ will have a coarser topology than the quotient topology: $\tau_Y\subseteq\tau_Z$.
But the identity $Y\to Z$ is also continuous by (2), so $Y$ has a finer topology: $\tau_Z\subseteq\tau_Y$.
So, $\tau_Y=\tau_Z$, i.e. as topological spaces, $Y=Z$. $\quad\quad$ -QED-