Show that $f(A)\cap B = f(A\cap f ^{−1} (B)).$ Let $f: X\to Y$ be a function and $A\subseteq X, B\subseteq Y$. Show that $f(A)\cap B = f\left(A\cap f^{−1}(B)\right)$
My work:
Suppose that $x\in f[A]\cap B$, such that there is an $a\in A$ in a way that $f(a)=x, x\in B$.
Since $f(a)\in B$, than $a\in A\cap f^{−1}(B)$, therefore $f(a)\in f[A \cap f^{−1}[B]]$, and: $f[A]\cap B \subseteq f[A \cap f^{−1}[B]].$
 A: Your first direction is great.
For the other direction,
if $y \in f(A \cap f^{-1}(B))$, then there is $ x \in X$ s.t $f(x)=y$ and $x\in A \cap f^{-1}(B) $ . Then by the second part, $f(x) \in B$ , and by the first part $y \in f(A)$
A: A different proof, using the fact that subsets form a Heyting algebra, $f^{-1}$ preserves the Heyting implication, and the adjuction $f \dashv f^{-1}$:
Given an arbitrary subset $C \subseteq Y$, we have
$\begin{equation}
\begin{split}
f(A) \cap B \subseteq C &\iff f(A) \subseteq (B \implies C) \\
&\iff A \subseteq f^{-1}(B \implies C) \\
&\iff A \subseteq (f^{-1}(B) \implies f^{-1}(C)) \\
&\iff A \cap f^{-1}(B) \subseteq f^{-1}(C) \\
&\iff f(A \cap f^{-1}(B)) \subseteq C
\end{split}
\end{equation}$
Therefore, we have $f(A) \cap B = f(A \cap f^{-1}(B))$ by the Yoneda Lemma. This proof holds in an arbitrary Heyting category (though typically, in category theory we write $\exists_f$ instead of $f$ to denote images).
For those who are not familiar with the notation $K \implies M$ where $K, M$ are sets, here is the definition. Given subsets $K, M \subseteq N$, we define $K \implies M$ to be the set $\{x \in N | x \in K \implies x \in M\}$. Note that given subsets $J, K, M \subseteq N$, we have $J \subseteq (K \implies M)$ iff $J \cap K \subseteq M$ (this is the property used above). And note that $f^{-1}(K \implies M) = (f^{-1}(K) \implies f^{-1}(M))$ (also used above).
