Solution
= Solution
\b[False.] Take $A=\{0,1\}$, $B=\{0\}$ and the <constant function> $f(a)=0$. For the <subsets> $X=\{0\}$ and $Y=\{1\}$, the <intersection> $X\cap Y$ is empty, so $f(X\cap Y)=\varnothing$, whereas the <intersection> of their <images of a function> is $f(X)\cap f(Y)=\{0\}$. In general the inclusion $f(X\cap Y)\subseteq f(X)\cap f(Y)$ holds, but a common image value can arise from different points of $X$ and $Y$.