[proofplan] The [contravariant Hom functor](/theorems/3973) is always left exact: surjectivity of $\pi$ gives injectivity of precomposition by $\pi$, and exactness at $B$ lets every map $B \to I$ vanishing on $A$ factor uniquely through $C$. Thus the only extra condition needed for short exactness is surjectivity of $\iota^*$, which says exactly that every $R$-[linear map](/page/Linear%20Map) $A \to I$ extends across every monomorphism $A \hookrightarrow B$. This is precisely the defining extension property of an injective module, and the converse is obtained by applying the assumed Hom exactness to the quotient sequence associated to an arbitrary monomorphism. [/proofplan] [step:Show that precomposition by $\pi$ is injective] Let \begin{align*} f_1, f_2: C \to I \end{align*} be $R$-linear maps such that $\pi^*(f_1)=\pi^*(f_2)$. This means \begin{align*} f_1 \circ \pi = f_2 \circ \pi. \end{align*} Because the sequence \begin{align*} 0 \longrightarrow A \xrightarrow{\iota} B \xrightarrow{\pi} C \longrightarrow 0 \end{align*} is exact, $\pi$ is surjective. Hence for every $c \in C$ there exists $b \in B$ such that $\pi(b)=c$. Therefore \begin{align*} f_1(c) = f_1(\pi(b)) = f_2(\pi(b)) = f_2(c). \end{align*} Thus $f_1=f_2$, so \begin{align*} \pi^*: \operatorname{Hom}_R(C,I) \to \operatorname{Hom}_R(B,I) \end{align*} is injective. [/step] [step:Identify the kernel of $\iota^*$ with the image of $\pi^*$] First, for every $R$-linear map \begin{align*} f: C \to I, \end{align*} we have \begin{align*} \iota^*(\pi^*(f)) = (f \circ \pi)\circ \iota = f \circ (\pi \circ \iota) = 0, \end{align*} because exactness gives $\pi \circ \iota=0$. Hence \begin{align*} \operatorname{im} \pi^* \subseteq \ker \iota^*. \end{align*} Conversely, let \begin{align*} g: B \to I \end{align*} be an $R$-linear map such that $g \in \ker \iota^*$. Then \begin{align*} g \circ \iota = 0. \end{align*} Define a map \begin{align*} h: C &\to I \end{align*} as follows. For $c \in C$, choose $b \in B$ with $\pi(b)=c$, and set \begin{align*} h(c) := g(b). \end{align*} This is well-defined: if $b_1,b_2 \in B$ satisfy $\pi(b_1)=\pi(b_2)$, then \begin{align*} \pi(b_1-b_2)=0, \end{align*} so $b_1-b_2 \in \ker \pi$. Exactness gives $\ker \pi=\operatorname{im}\iota$, so there exists $a \in A$ with \begin{align*} b_1-b_2=\iota(a). \end{align*} Therefore \begin{align*} g(b_1)-g(b_2)=g(b_1-b_2)=g(\iota(a))=0, \end{align*} and hence $g(b_1)=g(b_2)$. The map $h$ is $R$-linear. Indeed, for $c_1,c_2 \in C$, choose $b_1,b_2 \in B$ with $\pi(b_1)=c_1$ and $\pi(b_2)=c_2$. Then $\pi(b_1+b_2)=c_1+c_2$, so \begin{align*} h(c_1+c_2)=g(b_1+b_2)=g(b_1)+g(b_2)=h(c_1)+h(c_2). \end{align*} For $r \in R$ and $c \in C$, choose $b \in B$ with $\pi(b)=c$. Since $\pi(rb)=r\pi(b)=rc$, \begin{align*} h(rc)=g(rb)=r g(b)=r h(c). \end{align*} Thus \begin{align*} h \in \operatorname{Hom}_R(C,I). \end{align*} For every $b \in B$, \begin{align*} (\pi^*h)(b)=h(\pi(b))=g(b), \end{align*} so $g=\pi^*(h)$. Hence \begin{align*} \ker \iota^* \subseteq \operatorname{im}\pi^*. \end{align*} Combining the two inclusions gives \begin{align*} \ker \iota^*=\operatorname{im}\pi^*. \end{align*} [guided] The goal is to prove exactness at $\operatorname{Hom}_R(B,I)$. Exactness there means that the maps $g:B\to I$ killed by precomposition with $\iota$ are exactly the maps that come from maps $C\to I$ by precomposition with $\pi$. First take an $R$-linear map \begin{align*} f: C \to I. \end{align*} Then $\pi^*(f)=f\circ \pi$. Precomposing once more with $\iota$ gives \begin{align*} \iota^*(\pi^*(f)) = (f \circ \pi)\circ \iota = f \circ (\pi \circ \iota). \end{align*} The original sequence is exact, so $\operatorname{im}\iota=\ker\pi$, and in particular $\pi\circ\iota=0$. Therefore \begin{align*} \iota^*(\pi^*(f))=0. \end{align*} This proves \begin{align*} \operatorname{im}\pi^* \subseteq \ker\iota^*. \end{align*} Now suppose \begin{align*} g: B \to I \end{align*} is $R$-linear and lies in $\ker\iota^*$. By definition of $\iota^*$, this says \begin{align*} g\circ\iota=0. \end{align*} We want to construct an $R$-linear map $h:C\to I$ such that $g=h\circ\pi$. Since $\pi$ is surjective, every element $c\in C$ has the form $c=\pi(b)$ for some $b\in B$. This suggests defining \begin{align*} h(c):=g(b). \end{align*} The only issue is whether this depends on the chosen lift $b$. Let $b_1,b_2\in B$ satisfy $\pi(b_1)=\pi(b_2)$. Then \begin{align*} \pi(b_1-b_2)=0, \end{align*} so $b_1-b_2\in\ker\pi$. Exactness gives $\ker\pi=\operatorname{im}\iota$, hence there exists $a\in A$ such that \begin{align*} b_1-b_2=\iota(a). \end{align*} Because $g\circ\iota=0$, \begin{align*} g(b_1)-g(b_2)=g(b_1-b_2)=g(\iota(a))=0. \end{align*} Thus $g(b_1)=g(b_2)$, so the definition of $h$ is independent of the chosen lift. We next verify that $h$ is $R$-linear. Let $c_1,c_2\in C$, and choose $b_1,b_2\in B$ with $\pi(b_1)=c_1$ and $\pi(b_2)=c_2$. Since $\pi$ is $R$-linear, \begin{align*} \pi(b_1+b_2)=c_1+c_2. \end{align*} Therefore \begin{align*} h(c_1+c_2)=g(b_1+b_2)=g(b_1)+g(b_2)=h(c_1)+h(c_2). \end{align*} Similarly, for $r\in R$ and $c\in C$, choose $b\in B$ with $\pi(b)=c$. Then $\pi(rb)=rc$, and hence \begin{align*} h(rc)=g(rb)=r g(b)=r h(c). \end{align*} So $h\in\operatorname{Hom}_R(C,I)$. Finally, for every $b\in B$, \begin{align*} (\pi^*h)(b)=h(\pi(b))=g(b), \end{align*} where the last equality uses $b$ itself as a lift of $\pi(b)$. Thus $g=\pi^*(h)$, proving \begin{align*} \ker\iota^* \subseteq \operatorname{im}\pi^*. \end{align*} Together with the first inclusion, this gives \begin{align*} \ker\iota^*=\operatorname{im}\pi^*. \end{align*} [/guided] [/step] [step:Use injectivity of $I$ to prove surjectivity of $\iota^*$] Assume that $I$ is injective. Let \begin{align*} \varphi: A \to I \end{align*} be an $R$-linear map. Since the original sequence is exact, $\iota:A\to B$ is injective. By injectivity of $I$, applied to the monomorphism $\iota:A\to B$ and the map $\varphi:A\to I$, there exists an $R$-linear map \begin{align*} \Phi: B \to I \end{align*} such that \begin{align*} \Phi\circ\iota=\varphi. \end{align*} Equivalently, \begin{align*} \iota^*(\Phi)=\varphi. \end{align*} Thus $\iota^*$ is surjective. Together with the previous two steps, the induced Hom sequence is short exact. [/step] [step:Deduce injectivity of $I$ from exactness of every induced Hom sequence] Conversely, assume that for every short exact sequence of left $R$-modules \begin{align*} 0 \longrightarrow A \xrightarrow{\iota} B \xrightarrow{\pi} C \longrightarrow 0, \end{align*} the induced sequence \begin{align*} 0 \longrightarrow \operatorname{Hom}_R(C,I) \xrightarrow{\pi^*} \operatorname{Hom}_R(B,I) \xrightarrow{\iota^*} \operatorname{Hom}_R(A,I) \longrightarrow 0 \end{align*} is short exact. To prove that $I$ is injective, let \begin{align*} \alpha: M \to N \end{align*} be an injective $R$-linear map of left $R$-modules, and let \begin{align*} \psi: M \to I \end{align*} be an $R$-linear map. Define the quotient left $R$-module \begin{align*} Q := N/\alpha(M), \end{align*} and let \begin{align*} q: N \to Q \end{align*} be the quotient map. Since $\alpha$ is injective, $M$ identifies with the submodule $\alpha(M)\subset N$ through $\alpha$, and the sequence \begin{align*} 0 \longrightarrow M \xrightarrow{\alpha} N \xrightarrow{q} Q \longrightarrow 0 \end{align*} is exact. By the assumed exactness of the induced Hom sequence, the map \begin{align*} \alpha^*: \operatorname{Hom}_R(N,I) \to \operatorname{Hom}_R(M,I) \end{align*} is surjective. Hence there exists an $R$-linear map \begin{align*} \Psi: N \to I \end{align*} such that \begin{align*} \alpha^*(\Psi)=\psi. \end{align*} By definition of $\alpha^*$, this means \begin{align*} \Psi\circ\alpha=\psi. \end{align*} Therefore every $R$-linear map from a submodule domain extends across every injective $R$-linear map into $I$, so $I$ is injective. [guided] We now prove the converse direction directly from the definition of injectivity. The definition asks for the following extension property: whenever \begin{align*} \alpha: M \to N \end{align*} is an injective $R$-linear map and \begin{align*} \psi: M \to I \end{align*} is $R$-linear, there must exist an $R$-linear map \begin{align*} \Psi: N \to I \end{align*} such that $\Psi\circ\alpha=\psi$. So fix such $\alpha$ and $\psi$. To use the hypothesis, we must place $\alpha$ into a short exact sequence. Define \begin{align*} Q:=N/\alpha(M), \end{align*} and let \begin{align*} q:N\to Q \end{align*} be the quotient map. Because $\alpha$ is injective, its image $\alpha(M)$ is a submodule of $N$ isomorphic to $M$ through $\alpha$. The quotient map $q$ is surjective by construction, and its kernel is exactly $\alpha(M)$. Hence \begin{align*} 0 \longrightarrow M \xrightarrow{\alpha} N \xrightarrow{q} Q \longrightarrow 0 \end{align*} is a short exact sequence. Applying the assumed Hom exactness to this sequence gives a short exact sequence \begin{align*} 0 \longrightarrow \operatorname{Hom}_R(Q,I) \xrightarrow{q^*} \operatorname{Hom}_R(N,I) \xrightarrow{\alpha^*} \operatorname{Hom}_R(M,I) \longrightarrow 0. \end{align*} Exactness at the final term means precisely that \begin{align*} \alpha^*: \operatorname{Hom}_R(N,I) \to \operatorname{Hom}_R(M,I) \end{align*} is surjective. Since $\psi\in\operatorname{Hom}_R(M,I)$, there exists \begin{align*} \Psi\in\operatorname{Hom}_R(N,I) \end{align*} such that \begin{align*} \alpha^*(\Psi)=\psi. \end{align*} But $\alpha^*$ is precomposition with $\alpha$, so this equality is exactly \begin{align*} \Psi\circ\alpha=\psi. \end{align*} This is the required extension property. Therefore $I$ is injective. [/guided] [/step]