Let $X$ and $Y$ be sets and let $f\colon X\to Y$ be a function from $X$ to $Y$. If $A$ and $B$ are subsets of $X$, is it true that
$f(A)\cap f(B)$ is a subset of $f(A\cap B)$?
If so, prove your answer; otherwise, provide a counterexample.
If we assume that $f$ is injective, is the above inclusion true?
If we assume that $f$ is surjective, is the above inclusion true?
If we assume that $f$ is bijective, is the above inclusion true?
My thoughts:
No, it is not true. Counterexample, consider sets $A = \{1,2\}$ and $B = \{2,3\}$. Let $f\colon A\to B$ where $1$ maps to $4$, $2$ to nothing, and $3$ maps to $4$ and $5$.
Then $f(\{1,2\})\cap f(\{2,3\}) = \{4\}\cap\{4,5\} = \{4\}$. However, $f(\{1,2\}\cap\{2,3\}) = f(\{2\}) = \varnothing$. $\{4\}$ is not a subset of the empty set.
I think it's true if the function is injective (and also bijective obviously) but not surjective.
It looks like the converse is true. Since If $x$ is in both $A$ and $B$, then $f(x)$ is in both $f(A)$ and $f(B)$ so this means $f(A \cap B)$ is a subset of $f(A) \cap f(B)$.
To get an example that $f(A\cap B)\neq f(A)\cap f(B)$, Take $A = (-1, 0), B = (0, 1)$. Let $f(x) = x^2$. So $f(A \cap B)=f(\{0\})=0$, but $f(A) = f(B) = \{0, 1\}$ .