[proofplan] We prove the inclusion by taking an arbitrary element of $f(S \cap T)$ and unpacking the definition of image. The equality statement requires only the reverse inclusion, and this is where injectivity is used: an element lying in both $f(S)$ and $f(T)$ may have two witnesses, one in $S$ and one in $T$, and injectivity forces those witnesses to be equal. The common witness then belongs to $S \cap T$, giving membership in $f(S \cap T)$. [/proofplan] [step:Show that every element of $f(S \cap T)$ lies in both $f(S)$ and $f(T)$] Let $y \in f(S \cap T)$. By the definition of the image of a subset under the function $f: A \to B$, there exists an element $x \in S \cap T$ such that $f(x)=y$. Since $x \in S \cap T$, the definition of intersection gives $x \in S$ and $x \in T$. Therefore $y=f(x) \in f(S)$ and $y=f(x) \in f(T)$. Hence $y \in f(S) \cap f(T)$. Since $y \in f(S \cap T)$ was arbitrary, this proves \begin{align*} f(S \cap T) \subset f(S) \cap f(T). \end{align*} [guided] We want to prove a subset relation, so we start with an arbitrary element of the left-hand side and show that it belongs to the right-hand side. Let $y \in f(S \cap T)$. The statement $y \in f(S \cap T)$ means, by definition of the image of a subset, that there is some element $x \in S \cap T$ with $f(x)=y$. Now we use the defining property of intersection. Since $x \in S \cap T$, the element $x$ belongs to both sets: \begin{align*} x \in S \end{align*} and \begin{align*} x \in T. \end{align*} Because the same element $x$ is in $S$ and satisfies $f(x)=y$, the definition of image gives $y \in f(S)$. Because the same element $x$ is also in $T$ and satisfies $f(x)=y$, the definition of image gives $y \in f(T)$. Therefore $y$ lies in the intersection: \begin{align*} y \in f(S) \cap f(T). \end{align*} We have shown that every arbitrary element $y$ of $f(S \cap T)$ lies in $f(S) \cap f(T)$, which is exactly the subset relation \begin{align*} f(S \cap T) \subset f(S) \cap f(T). \end{align*} [/guided] [/step] [step:Use injectivity to prove the reverse inclusion] Assume now that $f$ is injective. Let $y \in f(S) \cap f(T)$. Then $y \in f(S)$ and $y \in f(T)$. By the definition of image, there exist elements $s \in S$ and $t \in T$ such that \begin{align*} f(s)=y \end{align*} and \begin{align*} f(t)=y. \end{align*} Thus $f(s)=f(t)$. Since $f$ is injective, it follows that $s=t$. Because $s \in S$ and $t \in T$, the equality $s=t$ implies $s \in T$. Hence $s \in S \cap T$. Since $f(s)=y$, the definition of image gives $y \in f(S \cap T)$. Since $y \in f(S) \cap f(T)$ was arbitrary, this proves \begin{align*} f(S) \cap f(T) \subset f(S \cap T). \end{align*} [/step] [step:Combine the two inclusions to obtain equality] The first step gives \begin{align*} f(S \cap T) \subset f(S) \cap f(T). \end{align*} Under the additional hypothesis that $f$ is injective, the second step gives \begin{align*} f(S) \cap f(T) \subset f(S \cap T). \end{align*} Therefore, when $f$ is injective, \begin{align*} f(S \cap T) = f(S) \cap f(T). \end{align*} This proves both assertions. [/step]