In topological spaces, a function is defined to be continuous if for every open set, the preimage of $f$ is continuous, which leads to this question:
Let $f:(X, \mathcal{U}_X) \to (Y, \mathcal{U}_Y)$ be a function, what is the coarsest topology $(X, \mathcal{U}_X)$ topology where $f$ is continuous, since $f$ is chosen arbitrary; my guess this would be dependent on the topology $(Y, \mathcal{U}_Y)$, now we can choose:
$$\mathcal U_X = \{f^{-1}(\mathcal U_Y)~:~\mathcal U_Y \text{ is open in Y}\}$$
Now, $f^{-1}$ doesn't necessary need to cover $X$ so we have to choose $\mathcal U_X\cup X\cup \emptyset$, however I am unable to prove that this is indeed a topology, specifically proving that the intersection of finite open sets in $X$ is open, specifically this basically proves it, however it feels somewhat wrong
$$f^{-1}\Big(\bigcap \mathcal U_X \Big)=f^{-1}\Big(\bigcup \mathcal U_X^c \Big) = \bigcup f^{-1}\Big(\mathcal{U}_X^c\Big)=\bigcap f^{-1}\Big(\mathcal U_X\Big)$$
But intuitively this feels wrong... as this would also mean that any intersection of open sets is also open and that wrong.
This is an example of an induced topology. As $\emptyset=f^{-1}(\emptyset)$ and $X=f^{-1}(Y)$ one does not need exceptions for these cases. As $$f^{-1}\left(\bigcup_{i\in I}U_i\right)=\bigcup_{i\in I}f^{-1}( U_i)$$ and $$f^{-1}\left(\bigcap_{i\in I}U_i\right)=\bigcap_{i\in I}f^{-1}( U_i)$$ for all families of subsets $(U_i)_{i\in I}$ of $I$, then proving that the $f^{-1}(U)$ for open $U\subseteq Y$ form a topology on $X$ is straightforward.