[proofplan] We prove exactness after applying each Hom functor to an arbitrary short exact sequence. For the covariant functor, exactness follows from the fact that $i:A' \to A$ is the kernel of $p:A \to A''$: maps $X \to A$ killed by $p$ are exactly those that factor uniquely through $i$. For the contravariant functor, the dual argument uses that $p:A \to A''$ is the cokernel of $i:A' \to A$: maps $A \to X$ killed by precomposition with $i$ are exactly those that factor uniquely through $p$. [/proofplan] [step:Apply $\operatorname{Hom}_{\mathcal A}(X,-)$ to a short exact sequence] Let \begin{align*} 0 \longrightarrow A' \xrightarrow{i} A \xrightarrow{p} A'' \longrightarrow 0 \end{align*} be a short exact sequence in $\mathcal A$. By definition of short exactness in an abelian category, $i$ is a monomorphism, $p$ is an epimorphism, $i = \ker p$, and $p = \operatorname{coker} i$. Define group homomorphisms \begin{align*} i_*: \operatorname{Hom}_{\mathcal A}(X,A') &\longrightarrow \operatorname{Hom}_{\mathcal A}(X,A), & f &\longmapsto i \circ f, \\ p_*: \operatorname{Hom}_{\mathcal A}(X,A) &\longrightarrow \operatorname{Hom}_{\mathcal A}(X,A''), & g &\longmapsto p \circ g. \end{align*} Since composition in an abelian category is biadditive, both $i_*$ and $p_*$ are homomorphisms of abelian groups. Because $p \circ i = 0$, for every $f:X \to A'$ we have \begin{align*} (p_* \circ i_*)(f) = p \circ i \circ f = 0. \end{align*} Thus $\operatorname{im} i_* \subseteq \ker p_*$. [/step] [step:Use the kernel property of $i$ to identify $\ker p_*$ with $\operatorname{im} i_*$] We first prove that $i_*$ is injective. Let $f_1,f_2:X \to A'$ be morphisms such that $i_*(f_1)=i_*(f_2)$. Then \begin{align*} i \circ f_1 = i \circ f_2. \end{align*} Since $i$ is a monomorphism, it follows that $f_1=f_2$. Hence \begin{align*} 0 \longrightarrow \operatorname{Hom}_{\mathcal A}(X,A') \xrightarrow{i_*} \operatorname{Hom}_{\mathcal A}(X,A) \end{align*} is exact at $\operatorname{Hom}_{\mathcal A}(X,A')$. It remains to prove $\ker p_* \subseteq \operatorname{im} i_*$. Let $g:X \to A$ be a morphism with $g \in \ker p_*$. This means \begin{align*} p \circ g = 0. \end{align*} Since $i:A' \to A$ is the kernel of $p:A \to A''$, the universal property of the kernel gives a unique morphism $u:X \to A'$ such that \begin{align*} i \circ u = g. \end{align*} Therefore $g=i_*(u)$, so $g \in \operatorname{im} i_*$. Together with $\operatorname{im} i_* \subseteq \ker p_*$, this gives \begin{align*} \operatorname{im} i_* = \ker p_*. \end{align*} Thus \begin{align*} 0 \longrightarrow \operatorname{Hom}_{\mathcal A}(X,A') \xrightarrow{i_*} \operatorname{Hom}_{\mathcal A}(X,A) \xrightarrow{p_*} \operatorname{Hom}_{\mathcal A}(X,A'') \end{align*} is exact. [guided] We need to prove two exactness assertions for the covariant Hom sequence. First, exactness at $\operatorname{Hom}_{\mathcal A}(X,A')$ means that $i_*$ is injective. Suppose $f_1,f_2:X \to A'$ satisfy $i_*(f_1)=i_*(f_2)$. By definition of $i_*$, this says \begin{align*} i \circ f_1 = i \circ f_2. \end{align*} Because the original sequence is short exact, $i:A' \to A$ is a monomorphism. The defining cancellation property of a monomorphism therefore gives $f_1=f_2$. Hence $i_*$ is injective. Second, exactness at $\operatorname{Hom}_{\mathcal A}(X,A)$ means \begin{align*} \operatorname{im} i_* = \ker p_*. \end{align*} The inclusion $\operatorname{im} i_* \subseteq \ker p_*$ follows from the identity $p \circ i=0$: for every $f:X \to A'$, \begin{align*} p_*(i_*(f)) = p \circ i \circ f = 0. \end{align*} For the reverse inclusion, take a morphism $g:X \to A$ with $g \in \ker p_*$. By definition of $p_*$, this means \begin{align*} p \circ g = 0. \end{align*} The important structural fact is that $i:A' \to A$ is not merely any monomorphism: it is the kernel of $p:A \to A''$. The universal property of the kernel says precisely that every morphism into $A$ whose composite with $p$ is zero factors uniquely through $i$. Applying this to $g:X \to A$, there exists a unique morphism $u:X \to A'$ such that \begin{align*} i \circ u = g. \end{align*} Thus $g=i_*(u)$, so $g \in \operatorname{im} i_*$. Hence $\ker p_* \subseteq \operatorname{im} i_*$, and therefore \begin{align*} \operatorname{im} i_* = \ker p_*. \end{align*} This proves exactness of the covariant Hom sequence. [/guided] [/step] [step:Apply $\operatorname{Hom}_{\mathcal A}(-,X)$ and use the cokernel property of $p$] Define group homomorphisms \begin{align*} p^*: \operatorname{Hom}_{\mathcal A}(A'',X) &\longrightarrow \operatorname{Hom}_{\mathcal A}(A,X), & f &\longmapsto f \circ p, \\ i^*: \operatorname{Hom}_{\mathcal A}(A,X) &\longrightarrow \operatorname{Hom}_{\mathcal A}(A',X), & g &\longmapsto g \circ i. \end{align*} Again, biadditivity of composition makes these maps homomorphisms of abelian groups. Since $p \circ i=0$, for every $f:A'' \to X$ we have \begin{align*} (i^* \circ p^*)(f) = f \circ p \circ i = 0, \end{align*} so $\operatorname{im} p^* \subseteq \ker i^*$. The map $p^*$ is injective because $p$ is an epimorphism. Indeed, if $f_1,f_2:A'' \to X$ satisfy $p^*(f_1)=p^*(f_2)$, then \begin{align*} f_1 \circ p = f_2 \circ p, \end{align*} and the epimorphism property of $p$ gives $f_1=f_2$. Now let $g:A \to X$ satisfy $g \in \ker i^*$. Then \begin{align*} g \circ i = 0. \end{align*} Since $p:A \to A''$ is the cokernel of $i:A' \to A$, the universal property of the cokernel gives a unique morphism $u:A'' \to X$ such that \begin{align*} u \circ p = g. \end{align*} Thus $g=p^*(u)$, so $g \in \operatorname{im} p^*$. Therefore \begin{align*} \operatorname{im} p^* = \ker i^*. \end{align*} Hence \begin{align*} 0 \longrightarrow \operatorname{Hom}_{\mathcal A}(A'',X) \xrightarrow{p^*} \operatorname{Hom}_{\mathcal A}(A,X) \xrightarrow{i^*} \operatorname{Hom}_{\mathcal A}(A',X) \end{align*} is exact. [guided] For the [contravariant Hom functor](/theorems/3973), the arrows reverse. The map induced by $p:A \to A''$ is precomposition with $p$: \begin{align*} p^*: \operatorname{Hom}_{\mathcal A}(A'',X) &\longrightarrow \operatorname{Hom}_{\mathcal A}(A,X), & f &\longmapsto f \circ p. \end{align*} The map induced by $i:A' \to A$ is precomposition with $i$: \begin{align*} i^*: \operatorname{Hom}_{\mathcal A}(A,X) &\longrightarrow \operatorname{Hom}_{\mathcal A}(A',X), & g &\longmapsto g \circ i. \end{align*} Because composition in an abelian category is additive in each variable, both maps are homomorphisms of abelian groups. Exactness first requires $i^* \circ p^*=0$. For $f:A'' \to X$, \begin{align*} (i^* \circ p^*)(f) = i^*(f \circ p) = f \circ p \circ i. \end{align*} The original sequence is exact, so $p \circ i=0$. Hence \begin{align*} (i^* \circ p^*)(f)=0. \end{align*} This proves $\operatorname{im} p^* \subseteq \ker i^*$. Next, $p^*$ is injective because $p$ is an epimorphism. If $f_1,f_2:A'' \to X$ satisfy $p^*(f_1)=p^*(f_2)$, then \begin{align*} f_1 \circ p = f_2 \circ p. \end{align*} Since $p$ is an epimorphism, right cancellation along $p$ gives $f_1=f_2$. Thus the sequence is exact at $\operatorname{Hom}_{\mathcal A}(A'',X)$. Finally, we identify the kernel of $i^*$. Let $g:A \to X$ be a morphism with $g \in \ker i^*$. By definition, \begin{align*} g \circ i = 0. \end{align*} Here the relevant part of short exactness is that $p:A \to A''$ is the cokernel of $i:A' \to A$. The universal property of the cokernel says that every morphism out of $A$ which kills $i$ factors uniquely through $p$. Applying this to $g:A \to X$, there exists a unique morphism $u:A'' \to X$ such that \begin{align*} u \circ p = g. \end{align*} Therefore $g=p^*(u)$, so $g \in \operatorname{im} p^*$. This proves $\ker i^* \subseteq \operatorname{im} p^*$. Combined with the reverse inclusion already proved, we obtain \begin{align*} \operatorname{im} p^* = \ker i^*. \end{align*} Thus the contravariant Hom sequence is exact. [/guided] [/step] [step:Conclude both Hom functors are left exact] The covariant argument shows that $\operatorname{Hom}_{\mathcal A}(X,-)$ sends every short exact sequence in $\mathcal A$ to an exact sequence through the first three terms. Therefore $\operatorname{Hom}_{\mathcal A}(X,-)$ is left exact. The contravariant argument shows that $\operatorname{Hom}_{\mathcal A}(-,X)$ sends every short exact sequence \begin{align*} 0 \longrightarrow A' \xrightarrow{i} A \xrightarrow{p} A'' \longrightarrow 0 \end{align*} to the exact sequence \begin{align*} 0 \longrightarrow \operatorname{Hom}_{\mathcal A}(A'',X) \xrightarrow{p^*} \operatorname{Hom}_{\mathcal A}(A,X) \xrightarrow{i^*} \operatorname{Hom}_{\mathcal A}(A',X). \end{align*} Equivalently, as a functor $\mathcal A^{\operatorname{op}} \to \operatorname{Ab}$, $\operatorname{Hom}_{\mathcal A}(-,X)$ is left exact. This proves both assertions. [/step]