$f(S \cap T) \neq f(S) \cap f(T)$
but
$f^{-1}(S \cap T)$ = $f^{-1}(S) \cap f^{-1}(T) $
where $f^{-1}$ is a preimage
what is a preimage and what difference does it make?
$f(S \cap T) \neq f(S) \cap f(T)$
but
$f^{-1}(S \cap T)$ = $f^{-1}(S) \cap f^{-1}(T) $
where $f^{-1}$ is a preimage
what is a preimage and what difference does it make?
If you have a function $f$ from some set $X$ to some set $Y$, and you have a subset $S$ of $Y$, then the preimage of $S$ is all those $x$-values in $X$ such that $f(x)$ is in $S$. Now I suggest you make up some examples of functions and see for yourself what $f(S\cap T)$ and $f^{-1}(S\cap T)$ and the rest look like.
A pre-image is the set of all points that map to some point in the target set.
For example, if $f(x) = 1$ if $x$ is positive and even, and $f(x) = 0$ if $x$ is positive and odd, then the pre-image of $f(x)$ is set of all positive integers. Do you see why?
This is different from an inverse. An inverse, if it exists, will be such that $f^{-1}(f(x)) = x$. Unfortunately, we tend to re-use the notation $f^{-1}$. Remember that a function may not be invertible; however, every function has a pre-image.
Using this, can you solve the problem?