[proofplan] We prove the two assertions by diagram chasing. For injectivity, we start with an element of $\ker f_3$, push it right to use injectivity of $f_4$, pull it left by exactness, and then use surjectivity of $f_1$ together with injectivity of $f_2$ to show that the original element is zero. For surjectivity, we start with an arbitrary element of $B_3$, push it right, lift through $f_4$, use exactness and injectivity of $f_5$ to pull the lift back to $A_3$, and then correct the remaining error by surjectivity of $f_2$. [/proofplan]

[step:Prove injectivity of $f_3$ by chasing an element of its kernel]Assume that $f_1$ is surjective and that $f_2$ and $f_4$ are injective. Let $a_3 \in A_3$ satisfy $f_3(a_3)=0_{B_3}$. We prove that $a_3=0_{A_3}$. By commutativity of the square involving $d_3,e_3,f_3,f_4$, \begin{align*} f_4(d_3(a_3)) = e_3(f_3(a_3)) = e_3(0_{B_3}) = 0_{B_4}. \end{align*} Since $f_4$ is injective, $d_3(a_3)=0_{A_4}$. Exactness of the top row at $A_3$ gives \begin{align*} \ker d_3 = \operatorname{im} d_2, \end{align*} so there exists $a_2 \in A_2$ such that $d_2(a_2)=a_3$. Now commutativity of the square involving $d_2,e_2,f_2,f_3$ gives \begin{align*} e_2(f_2(a_2)) = f_3(d_2(a_2)) = f_3(a_3) = 0_{B_3}. \end{align*} Thus $f_2(a_2) \in \ker e_2$. Exactness of the bottom row at $B_2$ gives $\ker e_2=\operatorname{im} e_1$, so there exists $b_1 \in B_1$ such that \begin{align*} e_1(b_1)=f_2(a_2). \end{align*} Since $f_1$ is surjective, there exists $a_1 \in A_1$ such that $f_1(a_1)=b_1$. Using commutativity of the square involving $d_1,e_1,f_1,f_2$, \begin{align*} f_2(d_1(a_1)) = e_1(f_1(a_1)) = e_1(b_1) = f_2(a_2). \end{align*} Hence \begin{align*} f_2(a_2-d_1(a_1))=0_{B_2}. \end{align*} Since $f_2$ is injective, $a_2-d_1(a_1)=0_{A_2}$, so $a_2=d_1(a_1)$. Therefore \begin{align*} a_3=d_2(a_2)=d_2(d_1(a_1)). \end{align*} Exactness of the top row at $A_2$ implies $\operatorname{im} d_1=\ker d_2$, hence $d_2(d_1(a_1))=0_{A_3}$. Therefore $a_3=0_{A_3}$. Since every element of $\ker f_3$ is zero, $f_3$ is injective.[/step]

[guided]Assume that $f_1$ is surjective and that $f_2$ and $f_4$ are injective. To prove that $f_3$ is injective, it is enough to prove that its kernel is zero. Let $a_3 \in A_3$ be an element with $f_3(a_3)=0_{B_3}$. First push $a_3$ one step to the right. Commutativity of the square from $A_3$ to $A_4$ and from $B_3$ to $B_4$ means that applying $d_3$ and then $f_4$ gives the same result as applying $f_3$ and then $e_3$. Thus \begin{align*} f_4(d_3(a_3)) = e_3(f_3(a_3)) = e_3(0_{B_3}) = 0_{B_4}. \end{align*} Because $f_4$ is injective, the only element of $A_4$ mapped to $0_{B_4}$ is $0_{A_4}$. Hence $d_3(a_3)=0_{A_4}$. Now exactness of the top row at $A_3$ identifies the kernel of $d_3$ with the image of $d_2$: \begin{align*} \ker d_3=\operatorname{im} d_2. \end{align*} Since $d_3(a_3)=0_{A_4}$, we have $a_3 \in \ker d_3$, so there exists $a_2 \in A_2$ such that \begin{align*} d_2(a_2)=a_3. \end{align*} Next we compare $a_2$ with the image of $A_1$. Commutativity of the square from $A_2$ to $A_3$ and from $B_2$ to $B_3$ gives \begin{align*} e_2(f_2(a_2)) = f_3(d_2(a_2)) = f_3(a_3) = 0_{B_3}. \end{align*} Thus $f_2(a_2) \in \ker e_2$. Exactness of the bottom row at $B_2$ says \begin{align*} \ker e_2=\operatorname{im} e_1, \end{align*} so there exists $b_1 \in B_1$ such that \begin{align*} e_1(b_1)=f_2(a_2). \end{align*} The hypothesis that $f_1$ is surjective lets us lift this element $b_1$ to the top row: there exists $a_1 \in A_1$ such that \begin{align*} f_1(a_1)=b_1. \end{align*} We now use the left square to compare $a_2$ with $d_1(a_1)$. Commutativity gives \begin{align*} f_2(d_1(a_1)) = e_1(f_1(a_1)) = e_1(b_1) = f_2(a_2). \end{align*} Subtracting in the abelian group $B_2$, \begin{align*} f_2(a_2-d_1(a_1))=0_{B_2}. \end{align*} Since $f_2$ is injective, its kernel is zero. Therefore \begin{align*} a_2-d_1(a_1)=0_{A_2}, \end{align*} so $a_2=d_1(a_1)$. Finally, \begin{align*} a_3=d_2(a_2)=d_2(d_1(a_1)). \end{align*} Exactness of the top row at $A_2$ gives $\operatorname{im} d_1=\ker d_2$, so every element in the image of $d_1$ is killed by $d_2$. Hence \begin{align*} d_2(d_1(a_1))=0_{A_3}. \end{align*} Thus $a_3=0_{A_3}$. We have shown that every element of $\ker f_3$ is zero, so $f_3$ is injective.[/guided]

[step:Prove surjectivity of $f_3$ by lifting and correcting an arbitrary element of $B_3$]Assume that $f_2$ and $f_4$ are surjective and that $f_5$ is injective. Let $b_3 \in B_3$. We construct $a_3 \in A_3$ such that $f_3(a_3)=b_3$. Since $f_4$ is surjective, there exists $a_4 \in A_4$ such that \begin{align*} f_4(a_4)=e_3(b_3). \end{align*} Using commutativity of the square involving $d_4,e_4,f_4,f_5$ and exactness of the bottom row at $B_4$, \begin{align*} f_5(d_4(a_4)) = e_4(f_4(a_4)) = e_4(e_3(b_3)) = 0_{B_5}. \end{align*} Since $f_5$ is injective, $d_4(a_4)=0_{A_5}$. Exactness of the top row at $A_4$ gives $\ker d_4=\operatorname{im} d_3$, so there exists $a_3' \in A_3$ such that \begin{align*} d_3(a_3')=a_4. \end{align*} Now compare $b_3$ with $f_3(a_3')$. By commutativity of the square involving $d_3,e_3,f_3,f_4$, \begin{align*} e_3(f_3(a_3'))=f_4(d_3(a_3'))=f_4(a_4)=e_3(b_3). \end{align*} Thus \begin{align*} e_3(b_3-f_3(a_3'))=0_{B_4}. \end{align*} Therefore $b_3-f_3(a_3') \in \ker e_3$. Exactness of the bottom row at $B_3$ gives $\ker e_3=\operatorname{im} e_2$, so there exists $b_2 \in B_2$ such that \begin{align*} e_2(b_2)=b_3-f_3(a_3'). \end{align*} Since $f_2$ is surjective, there exists $a_2 \in A_2$ such that $f_2(a_2)=b_2$. Define $a_3 \in A_3$ by \begin{align*} a_3 := a_3' + d_2(a_2). \end{align*} Using commutativity of the square involving $d_2,e_2,f_2,f_3$, \begin{align*} f_3(a_3) &= f_3(a_3') + f_3(d_2(a_2)) \\ &= f_3(a_3') + e_2(f_2(a_2)) \\ &= f_3(a_3') + e_2(b_2) \\ &= f_3(a_3') + b_3 - f_3(a_3') \\ &= b_3. \end{align*} Since every $b_3 \in B_3$ has a preimage under $f_3$, the homomorphism $f_3$ is surjective.[/step]

[guided]Assume that $f_2$ and $f_4$ are surjective and that $f_5$ is injective. To prove that $f_3$ is surjective, take an arbitrary element $b_3 \in B_3$ and construct an element of $A_3$ mapping to it. The first move is to push $b_3$ one step to the right. The element $e_3(b_3)$ lies in $B_4$. Since $f_4:A_4 \to B_4$ is surjective, we can lift this element to the top row: there exists $a_4 \in A_4$ such that \begin{align*} f_4(a_4)=e_3(b_3). \end{align*} We need this lift $a_4$ to come from $A_3$. Exactness of the top row says that this will happen precisely if $a_4 \in \ker d_4$. We verify that condition by pushing $a_4$ one step farther right. By commutativity of the square from $A_4$ to $A_5$ and from $B_4$ to $B_5$, \begin{align*} f_5(d_4(a_4)) = e_4(f_4(a_4)). \end{align*} Using the choice of $a_4$ and exactness of the bottom row at $B_4$, we get \begin{align*} f_5(d_4(a_4)) = e_4(e_3(b_3)) = 0_{B_5}, \end{align*} because $\operatorname{im} e_3=\ker e_4$. Since $f_5$ is injective, its kernel is zero, so \begin{align*} d_4(a_4)=0_{A_5}. \end{align*} Thus $a_4 \in \ker d_4$. Exactness of the top row at $A_4$ gives \begin{align*} \ker d_4=\operatorname{im} d_3, \end{align*} so there exists $a_3' \in A_3$ such that \begin{align*} d_3(a_3')=a_4. \end{align*} The element $a_3'$ is a first approximation to a preimage of $b_3$. We now compute the error. By commutativity of the square from $A_3$ to $A_4$ and from $B_3$ to $B_4$, \begin{align*} e_3(f_3(a_3'))=f_4(d_3(a_3'))=f_4(a_4)=e_3(b_3). \end{align*} Subtracting in the abelian group $B_4$, \begin{align*} e_3(b_3-f_3(a_3'))=0_{B_4}. \end{align*} Hence the error $b_3-f_3(a_3')$ lies in $\ker e_3$. Exactness of the bottom row at $B_3$ identifies this kernel with the image of $e_2$: \begin{align*} \ker e_3=\operatorname{im} e_2. \end{align*} Therefore there exists $b_2 \in B_2$ such that \begin{align*} e_2(b_2)=b_3-f_3(a_3'). \end{align*} Since $f_2:A_2 \to B_2$ is surjective, there exists $a_2 \in A_2$ such that \begin{align*} f_2(a_2)=b_2. \end{align*} Now correct the first approximation $a_3'$ by adding the boundary $d_2(a_2)$. Define $a_3 \in A_3$ by \begin{align*} a_3 := a_3' + d_2(a_2). \end{align*} We compute its image under $f_3$. Since $f_3$ is a group homomorphism and the square involving $d_2,e_2,f_2,f_3$ commutes, \begin{align*} f_3(a_3) &= f_3(a_3') + f_3(d_2(a_2)) \\ &= f_3(a_3') + e_2(f_2(a_2)) \\ &= f_3(a_3') + e_2(b_2) \\ &= f_3(a_3') + b_3 - f_3(a_3') \\ &= b_3. \end{align*} Thus the arbitrary element $b_3 \in B_3$ has a preimage $a_3 \in A_3$ under $f_3$. Therefore $f_3$ is surjective.[/guided]