[proofplan] We prove the two assertions separately. For injectivity, take an element of $B$ killed by $f_B$, push it to $C$, and use injectivity of $f_C$ to force it into $\ker p = \operatorname{im} i$; then commutativity and injectivity of $f_A$ force the original element to vanish. For surjectivity, start with an element of $\widetilde{B}$, lift its image in $\widetilde{C}$ through $f_C$ and $p$, and then correct the remaining error in $\ker \widetilde{p} = \operatorname{im} \widetilde{i}$ using surjectivity of $f_A$. [/proofplan] [step:Prove injectivity of $f_B$ from injectivity of $f_A$ and $f_C$] Assume that $f_A$ and $f_C$ are injective. Let $b \in B$ satisfy $f_B(b) = 0_{\widetilde{B}}$. Applying $\widetilde{p}$ and using commutativity of the right square gives \begin{align*} 0_{\widetilde{C}} = \widetilde{p}(0_{\widetilde{B}}) = \widetilde{p}(f_B(b)) = f_C(p(b)). \end{align*} Since $f_C$ is injective, $p(b) = 0_C$. Exactness of the top row at $B$ gives $\ker p = \operatorname{im} i$, so there exists $a \in A$ such that $i(a) = b$. Using commutativity of the left square, \begin{align*} \widetilde{i}(f_A(a)) = f_B(i(a)) = f_B(b) = 0_{\widetilde{B}}. \end{align*} Exactness of the bottom row at $\widetilde{A}$ implies that $\widetilde{i}$ is injective, hence $f_A(a) = 0_{\widetilde{A}}$. Since $f_A$ is injective, $a = 0_A$. Therefore \begin{align*} b = i(a) = i(0_A) = 0_B. \end{align*} Thus $\ker f_B = \{0_B\}$, so $f_B$ is injective. [guided] Assume that $f_A$ and $f_C$ are injective. To prove that $f_B$ is injective, we must prove that the only element of $B$ mapped to $0_{\widetilde{B}}$ is $0_B$. Let $b \in B$ be such that $f_B(b) = 0_{\widetilde{B}}$. We first move $b$ to the right side of the diagram. Applying $\widetilde{p}: \widetilde{B} \to \widetilde{C}$ to the equality $f_B(b) = 0_{\widetilde{B}}$ and using the commutativity relation $\widetilde{p} \circ f_B = f_C \circ p$, we obtain \begin{align*} 0_{\widetilde{C}} = \widetilde{p}(0_{\widetilde{B}}) = \widetilde{p}(f_B(b)) = f_C(p(b)). \end{align*} Since $f_C$ is injective, the equality $f_C(p(b)) = 0_{\widetilde{C}}$ forces $p(b) = 0_C$. Exactness of the top row at $B$ says precisely that $\ker p = \operatorname{im} i$. Therefore there exists $a \in A$ such that \begin{align*} i(a) = b. \end{align*} Now we use the left square. The commutativity relation $f_B \circ i = \widetilde{i} \circ f_A$ gives \begin{align*} \widetilde{i}(f_A(a)) = f_B(i(a)) = f_B(b) = 0_{\widetilde{B}}. \end{align*} Exactness of the bottom row at $\widetilde{A}$ implies that $\widetilde{i}: \widetilde{A} \to \widetilde{B}$ is injective, because the preceding map into $\widetilde{A}$ is the zero map and hence $\ker \widetilde{i} = \{0_{\widetilde{A}}\}$. Thus $f_A(a) = 0_{\widetilde{A}}$. Since $f_A$ is injective, $a = 0_A$. Hence \begin{align*} b = i(a) = i(0_A) = 0_B. \end{align*} We have shown that every element of $\ker f_B$ is zero, so $f_B$ is injective. [/guided] [/step] [step:Prove surjectivity of $f_B$ from surjectivity of $f_A$ and $f_C$] Assume that $f_A$ and $f_C$ are surjective. Let $\widetilde{b} \in \widetilde{B}$. Since $f_C$ is surjective, there exists $c \in C$ such that \begin{align*} f_C(c) = \widetilde{p}(\widetilde{b}). \end{align*} Exactness of the top row at $C$ implies that $p$ is surjective, so there exists $b \in B$ such that $p(b) = c$. Define the error element $\widetilde{e} \in \widetilde{B}$ by \begin{align*} \widetilde{e} := \widetilde{b} - f_B(b). \end{align*} Using commutativity of the right square, \begin{align*} \widetilde{p}(\widetilde{e}) = \widetilde{p}(\widetilde{b}) - \widetilde{p}(f_B(b)) = f_C(c) - f_C(p(b)) = f_C(c) - f_C(c) = 0_{\widetilde{C}}. \end{align*} Thus $\widetilde{e} \in \ker \widetilde{p}$. Exactness of the bottom row at $\widetilde{B}$ gives $\ker \widetilde{p} = \operatorname{im} \widetilde{i}$, so there exists $\widetilde{a} \in \widetilde{A}$ such that $\widetilde{i}(\widetilde{a}) = \widetilde{e}$. Since $f_A$ is surjective, there exists $a \in A$ such that $f_A(a) = \widetilde{a}$. Then, using commutativity of the left square, \begin{align*} f_B(b + i(a)) = f_B(b) + f_B(i(a)) = f_B(b) + \widetilde{i}(f_A(a)) = f_B(b) + \widetilde{i}(\widetilde{a}) = f_B(b) + \widetilde{e} = \widetilde{b}. \end{align*} Therefore every $\widetilde{b} \in \widetilde{B}$ lies in the image of $f_B$, so $f_B$ is surjective. [guided] Assume that $f_A$ and $f_C$ are surjective. To prove that $f_B$ is surjective, we start with an arbitrary target element $\widetilde{b} \in \widetilde{B}$ and construct an element of $B$ that maps to it. First we match the image of $\widetilde{b}$ in $\widetilde{C}$. Since $f_C: C \to \widetilde{C}$ is surjective, there exists $c \in C$ such that \begin{align*} f_C(c) = \widetilde{p}(\widetilde{b}). \end{align*} Exactness of the top row at $C$ means that $p: B \to C$ is surjective. Hence there exists $b \in B$ such that \begin{align*} p(b) = c. \end{align*} The element $b$ has the correct image after passing all the way to $\widetilde{C}$, but $f_B(b)$ need not equal $\widetilde{b}$ in $\widetilde{B}$. We measure the difference by defining the error element $\widetilde{e} \in \widetilde{B}$ by \begin{align*} \widetilde{e} := \widetilde{b} - f_B(b). \end{align*} We now prove that this error lies in the image of $\widetilde{i}$. Applying $\widetilde{p}$ and using the commutativity relation $\widetilde{p} \circ f_B = f_C \circ p$, we get \begin{align*} \widetilde{p}(\widetilde{e}) = \widetilde{p}(\widetilde{b} - f_B(b)) = \widetilde{p}(\widetilde{b}) - \widetilde{p}(f_B(b)) = f_C(c) - f_C(p(b)) = f_C(c) - f_C(c) = 0_{\widetilde{C}}. \end{align*} Thus $\widetilde{e} \in \ker \widetilde{p}$. Exactness of the bottom row at $\widetilde{B}$ says that $\ker \widetilde{p} = \operatorname{im} \widetilde{i}$, so there exists $\widetilde{a} \in \widetilde{A}$ such that \begin{align*} \widetilde{i}(\widetilde{a}) = \widetilde{e}. \end{align*} It remains to lift this correction term back to $A$. Since $f_A: A \to \widetilde{A}$ is surjective, there exists $a \in A$ such that \begin{align*} f_A(a) = \widetilde{a}. \end{align*} Now add the correction $i(a)$ to $b$. Using that $f_B$ is an $R$-module homomorphism and using the commutativity relation $f_B \circ i = \widetilde{i} \circ f_A$, we compute \begin{align*} f_B(b + i(a)) = f_B(b) + f_B(i(a)) = f_B(b) + \widetilde{i}(f_A(a)) = f_B(b) + \widetilde{i}(\widetilde{a}) = f_B(b) + \widetilde{e} = f_B(b) + \widetilde{b} - f_B(b) = \widetilde{b}. \end{align*} Thus the arbitrary element $\widetilde{b} \in \widetilde{B}$ has a preimage $b + i(a) \in B$ under $f_B$. Therefore $f_B$ is surjective. [/guided] [/step]