A topology finer than the final topology? Suppose that $f: X \to Y$ is a function and $X$ has a topology on it. The final topology on $Y$ so that $f$ is continuous is the collection of open sets $U \in \mathscr{T}_Y$ such that $$ f^{-1}(U) \in \mathscr{T}_X \iff U \in \mathscr{T}_X $$ Equivalently the final topology on $Y$ is the image of all saturated open set in $X$. * * * My question is why not just push forward "unsaturated" open sets $U \in \mathscr{T}_X$ and declare $f(U)$ to be open? I can see why this might be problematic. For if $U$ is not saturated and we declare $f(U) = V$ to be open, then by definition of being "unsaturated", we have that if $H = f^{-1}(V)$ then $$ H \not = U$$ And so, whatever $H$ is, we don't know if it is open or not. This does not definitively rule out the possibility that $H$ is open, however. * * * TLDR: Is there a definitive reason why we don't push forward "unsaturated" sets.

Veridian Dynamics has motivated an answer to my almost trivial question:

This approach of pushing forward unsaturated open sets could work. That is, we could be successful in doing the following:

1. Select unsaturated $U$ and push it forward to obtain $V = f(U)$
2. Declare $V$ to be open.
3. Observe that $f^{-1}(V)$ (which is $\
ot = U$) is open
4. But (and here's the point) if $f^{-1}(V)$ is open, then we would have already declared $V$ to be open by definition of the final topology. Thus, if there is an instance when pushing forward an unsaturated set will give us a "valid" open set $V$ (by "valid" I mean that it will respect continuity of $f$), then $V$ has already been added to the collection as a result of pushing forward a _saturated_ set.