[proofplan]
We prove the two containments separately. First, any scalar that annihilates $B$ also annihilates $A$ by injectivity of $f$, and annihilates $C$ by surjectivity of $g$. Second, if one scalar annihilates $A$ and another annihilates $C$, then their product annihilates every element of $B$ because exactness forces the intermediate element into $\operatorname{im} f$. Finally, we pass from single products to the whole product ideal by taking finite sums.
[/proofplan]
[step:Show that every scalar annihilating $B$ annihilates both $A$ and $C$]
Let $r \in \operatorname{Ann}_R(B)$. We prove that $r \in \operatorname{Ann}_R(A)$ and $r \in \operatorname{Ann}_R(C)$.
Let $x \in A$. Since $f: A \to B$ is an $R$-module homomorphism,
\begin{align*}
f(rx)=r f(x).
\end{align*}
Because $r \in \operatorname{Ann}_R(B)$ and $f(x)\in B$, we have $r f(x)=0$. Hence $f(rx)=0$. Since exactness at $A$ says that $f$ is injective, it follows that $rx=0$. As $x \in A$ was arbitrary, $r \in \operatorname{Ann}_R(A)$.
Let $z \in C$. Since exactness at $C$ says that $g: B \to C$ is surjective, there exists $y \in B$ such that $g(y)=z$. Using $R$-linearity of $g$ and the assumption that $r$ annihilates $B$,
\begin{align*}
rz=r g(y)=g(ry)=g(0)=0.
\end{align*}
Thus $r \in \operatorname{Ann}_R(C)$. Therefore
\begin{align*}
\operatorname{Ann}_R(B) \subset \operatorname{Ann}_R(A)\cap \operatorname{Ann}_R(C).
\end{align*}
[guided]
Let $r \in \operatorname{Ann}_R(B)$. By definition, this means $ry=0$ for every $y \in B$. We must prove that the same scalar $r$ annihilates every element of $A$ and every element of $C$.
First take an arbitrary element $x \in A$. The map $f: A \to B$ is $R$-linear, so it commutes with scalar multiplication:
\begin{align*}
f(rx)=r f(x).
\end{align*}
The element $f(x)$ lies in $B$, and $r$ annihilates all of $B$, so $r f(x)=0$. Hence $f(rx)=0$. Exactness at $A$ means that $f$ is injective, so the only element of $A$ mapped to $0$ is $0$. Therefore $rx=0$. Since $x$ was arbitrary, $r$ annihilates $A$, so $r \in \operatorname{Ann}_R(A)$.
Now take an arbitrary element $z \in C$. We cannot directly compare $z$ with an element of $B$, so we use surjectivity of $g$. Exactness at $C$ says that $g: B \to C$ is surjective, hence there exists $y \in B$ with $g(y)=z$. Since $g$ is $R$-linear,
\begin{align*}
rz=r g(y)=g(ry).
\end{align*}
But $r$ annihilates $B$, so $ry=0$. Therefore
\begin{align*}
rz=g(0)=0.
\end{align*}
Thus $r$ annihilates every element of $C$, so $r \in \operatorname{Ann}_R(C)$.
We have shown that every $r \in \operatorname{Ann}_R(B)$ lies in both $\operatorname{Ann}_R(A)$ and $\operatorname{Ann}_R(C)$. Hence
\begin{align*}
\operatorname{Ann}_R(B) \subset \operatorname{Ann}_R(A)\cap \operatorname{Ann}_R(C).
\end{align*}
[/guided]
[/step]
[step:Show that each product of annihilators of $A$ and $C$ annihilates $B$]
Let $\alpha \in \operatorname{Ann}_R(A)$ and $\gamma \in \operatorname{Ann}_R(C)$. We prove that $\alpha\gamma \in \operatorname{Ann}_R(B)$.
Let $y \in B$. Since $g$ is $R$-linear,
\begin{align*}
g(\gamma y)=\gamma g(y).
\end{align*}
Because $g(y)\in C$ and $\gamma \in \operatorname{Ann}_R(C)$, we have $\gamma g(y)=0$. Hence $\gamma y \in \ker g$. Exactness at $B$ gives $\ker g=\operatorname{im} f$, so there exists $x \in A$ such that
\begin{align*}
f(x)=\gamma y.
\end{align*}
Using commutativity of $R$, $R$-linearity of $f$, and $\alpha \in \operatorname{Ann}_R(A)$, we obtain
\begin{align*}
(\alpha\gamma)y=\alpha(\gamma y)=\alpha f(x)=f(\alpha x)=f(0)=0.
\end{align*}
Thus $\alpha\gamma$ annihilates every $y \in B$, so $\alpha\gamma \in \operatorname{Ann}_R(B)$.
[/step]
[step:Pass from pure products to the product ideal]
By [citetheorem:7850], $\operatorname{Ann}_R(A)$, $\operatorname{Ann}_R(B)$, and $\operatorname{Ann}_R(C)$ are ideals of $R$. Let
\begin{align*}
t \in \operatorname{Ann}_R(A)\operatorname{Ann}_R(C).
\end{align*}
By definition of the product of ideals, there exist $n \in \mathbb{N}$, elements $\alpha_1,\dots,\alpha_n \in \operatorname{Ann}_R(A)$, and elements $\gamma_1,\dots,\gamma_n \in \operatorname{Ann}_R(C)$ such that
\begin{align*}
t=\sum_{i=1}^{n}\alpha_i\gamma_i.
\end{align*}
The previous step shows that $\alpha_i\gamma_i \in \operatorname{Ann}_R(B)$ for each $i \in \{1,\dots,n\}$. Since $\operatorname{Ann}_R(B)$ is an ideal, it is closed under finite sums. Therefore $t \in \operatorname{Ann}_R(B)$. Hence
\begin{align*}
\operatorname{Ann}_R(A)\operatorname{Ann}_R(C) \subset \operatorname{Ann}_R(B).
\end{align*}
Combining this with the right containment proved above gives
\begin{align*}
\operatorname{Ann}_R(A)\operatorname{Ann}_R(C) \subset \operatorname{Ann}_R(B) \subset \operatorname{Ann}_R(A)\cap \operatorname{Ann}_R(C).
\end{align*}
This proves the theorem.
[/step]
