[proofplan] The [contravariant Hom functor](/theorems/3973) is always left exact: a short exact sequence $0 \to A \to B \to C \to 0$ gives exactness at $\operatorname{Hom}_{\mathcal A}(C,I)$ and $\operatorname{Hom}_{\mathcal A}(B,I)$. Thus exactness is equivalent to the remaining surjectivity of the restriction map $\operatorname{Hom}_{\mathcal A}(B,I)\to \operatorname{Hom}_{\mathcal A}(A,I)$ for every monomorphism $A\to B$. That surjectivity is precisely the extension property defining injectivity of $I$. [/proofplan] [step:Define the induced Hom maps for a short exact sequence] Let \begin{align*} 0 \longrightarrow A \xrightarrow{u} B \xrightarrow{v} C \longrightarrow 0 \end{align*} be a short exact sequence in $\mathcal A$. Define group homomorphisms \begin{align*} v^*: \operatorname{Hom}_{\mathcal A}(C,I) &\to \operatorname{Hom}_{\mathcal A}(B,I),& \varphi &\mapsto \varphi \circ v, \\ u^*: \operatorname{Hom}_{\mathcal A}(B,I) &\to \operatorname{Hom}_{\mathcal A}(A,I),& \psi &\mapsto \psi \circ u. \end{align*} Since $v\circ u=0$, for every $\varphi:A\to I$ one has \begin{align*} u^*(v^*(\varphi))=(\varphi\circ v)\circ u=\varphi\circ (v\circ u)=0. \end{align*} Thus $\operatorname{im}(v^*)\subseteq \ker(u^*)$. [/step] [step:Prove the left exact part of the contravariant Hom sequence] We first prove that \begin{align*} 0 \longrightarrow \operatorname{Hom}_{\mathcal A}(C,I) \xrightarrow{v^*} \operatorname{Hom}_{\mathcal A}(B,I) \xrightarrow{u^*} \operatorname{Hom}_{\mathcal A}(A,I) \end{align*} is exact. Since $v$ is an epimorphism, if $\varphi_1,\varphi_2:C\to I$ satisfy $v^*(\varphi_1)=v^*(\varphi_2)$, then \begin{align*} \varphi_1\circ v=\varphi_2\circ v, \end{align*} and therefore $\varphi_1=\varphi_2$. Hence $v^*$ is injective. Now let $\psi:B\to I$ satisfy $u^*(\psi)=0$, so $\psi\circ u=0$. Because $v:B\to C$ is the cokernel of $u:A\to B$, the universal property of the cokernel gives a unique morphism $\varphi:C\to I$ such that \begin{align*} \varphi\circ v=\psi. \end{align*} Thus $\psi=v^*(\varphi)$, so $\ker(u^*)\subseteq \operatorname{im}(v^*)$. Together with the previous inclusion, this gives \begin{align*} \ker(u^*)=\operatorname{im}(v^*). \end{align*} [guided] We verify the part of exactness that does not use injectivity of $I$. The map $v^*$ is precomposition with $v$, so it sends a morphism $\varphi:C\to I$ to the composite $\varphi\circ v:B\to I$. To prove that $v^*$ is injective, suppose $\varphi_1,\varphi_2:C\to I$ satisfy \begin{align*} v^*(\varphi_1)=v^*(\varphi_2). \end{align*} By definition of $v^*$, this means \begin{align*} \varphi_1\circ v=\varphi_2\circ v. \end{align*} Since $v$ is an epimorphism in the short exact sequence, equality after precomposition with $v$ forces $\varphi_1=\varphi_2$. Hence $v^*$ is injective. Next we identify the kernel of $u^*$. Let $\psi:B\to I$ be a morphism with $u^*(\psi)=0$. This means \begin{align*} \psi\circ u=0. \end{align*} In the short exact sequence, $v:B\to C$ is a cokernel of $u:A\to B$. The universal property of this cokernel says that every morphism out of $B$ which kills $u$ factors uniquely through $v$. Applying this to $\psi:B\to I$, there is a unique morphism $\varphi:C\to I$ satisfying \begin{align*} \varphi\circ v=\psi. \end{align*} Therefore $\psi=v^*(\varphi)$, so every element of $\ker(u^*)$ lies in $\operatorname{im}(v^*)$. The reverse inclusion follows from $v\circ u=0$, because for every $\varphi:C\to I$, \begin{align*} u^*(v^*(\varphi))=(\varphi\circ v)\circ u=\varphi\circ (v\circ u)=0. \end{align*} Thus $\ker(u^*)=\operatorname{im}(v^*)$. [/guided] [/step] [step:Use injectivity of $I$ to prove exactness of the Hom functor] Assume that $I$ is injective. To prove that $\operatorname{Hom}_{\mathcal A}(-,I)$ is exact, it remains only to prove that $u^*$ is surjective for every short exact sequence as above. Let $f:A\to I$ be any morphism. Since $u:A\to B$ is a monomorphism and $I$ is injective, there exists a morphism $g:B\to I$ such that \begin{align*} g\circ u=f. \end{align*} By definition of $u^*$, this says $u^*(g)=f$. Hence $u^*$ is surjective. Combining this surjectivity with the left exactness proved above, the sequence \begin{align*} 0 \longrightarrow \operatorname{Hom}_{\mathcal A}(C,I) \xrightarrow{v^*} \operatorname{Hom}_{\mathcal A}(B,I) \xrightarrow{u^*} \operatorname{Hom}_{\mathcal A}(A,I) \longrightarrow 0 \end{align*} is exact for every short exact sequence in $\mathcal A$. Therefore $\operatorname{Hom}_{\mathcal A}(-,I):\mathcal A^{\operatorname{op}}\to\operatorname{Ab}$ is exact. [/step] [step:Use exactness of the Hom functor to recover the injective extension property] Conversely, assume that \begin{align*} \operatorname{Hom}_{\mathcal A}(-,I):\mathcal A^{\operatorname{op}}\to\operatorname{Ab} \end{align*} is exact. We prove that $I$ is injective. Let $u:A\to B$ be a monomorphism in $\mathcal A$, and let $f:A\to I$ be any morphism. Since $\mathcal A$ is abelian, $u$ has a cokernel. Let \begin{align*} v:B\to C \end{align*} be a cokernel of $u$. Because $u$ is a monomorphism and $v$ is its cokernel, the sequence \begin{align*} 0 \longrightarrow A \xrightarrow{u} B \xrightarrow{v} C \longrightarrow 0 \end{align*} is short exact. Exactness of $\operatorname{Hom}_{\mathcal A}(-,I)$ therefore gives exactness of \begin{align*} 0 \longrightarrow \operatorname{Hom}_{\mathcal A}(C,I) \xrightarrow{v^*} \operatorname{Hom}_{\mathcal A}(B,I) \xrightarrow{u^*} \operatorname{Hom}_{\mathcal A}(A,I) \longrightarrow 0. \end{align*} In particular, $u^*$ is surjective. Hence there exists a morphism $g:B\to I$ such that \begin{align*} u^*(g)=f. \end{align*} By definition of $u^*$, this means \begin{align*} g\circ u=f. \end{align*} Thus every morphism $f:A\to I$ extends along every monomorphism $u:A\to B$. This is exactly the injectivity of $I$. [/step] [step:Conclude the equivalence] We have shown that if $I$ is injective, then the contravariant Hom functor $\operatorname{Hom}_{\mathcal A}(-,I)$ sends every short exact sequence in $\mathcal A$ to a short exact sequence of abelian groups in the reversed direction. We have also shown that if this Hom functor is exact, then every morphism into $I$ extends along every monomorphism. Therefore $I$ is injective if and only if $\operatorname{Hom}_{\mathcal A}(-,I):\mathcal A^{\operatorname{op}}\to\operatorname{Ab}$ is exact. [/step]