Image of an intersection

Statement

Let $f \colon A \to B$ be a function and let $S$ and $T$ be subsets of $A$. Then the image of $S \cap T$ is a subset of the intersection of the images: $$f(S \cap T) \subseteq f(S) \cap f(T).$$

Remarks

The inclusion can be strict. For the constant function $f \colon \{1, 2\} \to \{0\}$, $S = \{1\}$ and $T = \{2\}$, the left side is $\varnothing$ and the right side is $\{0\}$. Equality holds for all $S$ and $T$ if and only if $f$ is injective. For unions, equality always holds: $f(S \cup T) = f(S) \cup f(T)$.

Explanations

L1 human

If $b = f(x)$ with $x$ in both $S$ and $T$, then $b$ is in both images. But $b$ can be in both images for another reason: $b = f(s) = f(t)$ with $s \in S$, $t \in T$ and $s \neq t$.