In this article, we present and prove a fundamental result from module theory which gives a characterization of exact sequences using the Hom functor. This theorem is central to homological algebra and helps in understanding how algebraic structures behave under homomorphisms.

Proposition

Let

\( M' \xrightarrow{u} M \xrightarrow{v} M'' \rightarrow 0 \)

be a sequence of \( A \)-modules and \( A \)-module homomorphisms.

Then the above sequence is exact if and only if for every \( A \)-module \( N \), the induced sequence

\( 0 \rightarrow \operatorname{Hom}(M'',N) \xrightarrow{\bar{v}} \operatorname{Hom}(M,N) \xrightarrow{\bar{u}} \operatorname{Hom}(M',N) \)

is exact, where the induced maps are defined by

\( \bar{v}(f) = f \circ v \),    \( \bar{u}(g) = g \circ u \).

Proof

Forward Direction (⇒)

Assume that the sequence

\( M' \xrightarrow{u} M \xrightarrow{v} M'' \rightarrow 0 \)

is exact. Then:

We show that the sequence

\( 0 \rightarrow \operatorname{Hom}(M'',N) \xrightarrow{\bar{v}} \operatorname{Hom}(M,N) \xrightarrow{\bar{u}} \operatorname{Hom}(M',N) \)

is exact.

Claim 1: \( \bar{v} \) is injective.

Suppose \( \bar{v}(f) = 0 \) for some \( f : M'' \to N \). Then

\( f \circ v = 0 \Rightarrow f(v(M)) = 0 \Rightarrow f(M'') = 0 \Rightarrow f = 0. \)

Hence, \( \bar{v} \) is injective.

Claim 2: \( \operatorname{Im}(\bar{v}) = \ker(\bar{u}) \).

Let \( f : M'' \to N \). Then

\( \bar{u}(\bar{v}(f)) = \bar{u}(f \circ v) = f \circ v \circ u = 0 \),

since \( v \circ u = 0 \). Therefore,

\( \operatorname{Im}(\bar{v}) \subseteq \ker(\bar{u}). \)

Now let \( g : M \to N \) such that \( g \in \ker(\bar{u}) \), i.e., \( g \circ u = 0 \).

We must show that \( g \in \operatorname{Im}(\bar{v}) \).

Since \( v \) is surjective, for any \( x \in M'' \), choose \( y \in M \) such that \( v(y) = x \).

Define \( h : M'' \to N \) by

\( h(x) = g(y). \)

This definition is well-defined: if \( y_1, y_2 \in M \) satisfy \( v(y_1) = v(y_2) \),

then \( y_1 - y_2 \in \ker(v) = \operatorname{Im}(u) \), so \( y_1 - y_2 = u(z) \) for some \( z \).

Thus,

\( g(y_1) - g(y_2) = g(u(z)) = 0 \Rightarrow g(y_1) = g(y_2). \)

Hence, \( h \) is well-defined and is a homomorphism. Furthermore,

\( (h \circ v)(y) = h(v(y)) = g(y), \)

so \( g = h \circ v \), implying

\( g \in \operatorname{Im}(\bar{v}). \)

Therefore,

\( \operatorname{Im}(\bar{v}) = \ker(\bar{u}). \)

Converse Direction (⇐)

Now suppose that for every \( A \)-module \( N \), the sequence

\( 0 \rightarrow \operatorname{Hom}(M'',N) \xrightarrow{\bar{v}} \operatorname{Hom}(M,N) \xrightarrow{\bar{u}} \operatorname{Hom}(M',N) \)

is exact. We show that

Claim 1: \( v \) is surjective.

Let \( N = M'' / \operatorname{Im}(v) \) and define \( f : M'' \to N \) by

\( f(x) = x + \operatorname{Im}(v). \)

Then \( \bar{v}(f) = f \circ v = 0 \).

Since \( \bar{v} \) is injective, we obtain \( f = 0 \), which implies \( M'' = \operatorname{Im}(v) \).

Hence, \( v \) is surjective.

Claim 2: \( \operatorname{Im}(u) = \ker(v) \).

Since \( \bar{u} \circ \bar{v} = 0 \), for every homomorphism \( f : M'' \to N \),

\( f \circ v \circ u = 0. \)

Taking \( f = \operatorname{id}_{M''} \), we obtain

\( v \circ u = 0 \Rightarrow \operatorname{Im}(u) \subseteq \ker(v). \)

Now let \( N = M / \operatorname{Im}(u) \), and define \( \phi : M \to N \) by

\( \phi(x) = x + \operatorname{Im}(u). \)

Then \( \ker(\phi) = \operatorname{Im}(u) \), and \( \bar{u}(\phi) = 0 \).

Hence, \( \phi \in \operatorname{Im}(\bar{v}) \), so there exists \( \psi : M'' \to N \) such that \( \phi = \psi \circ v \).

This implies

\( \ker(v) \subseteq \operatorname{Im}(u). \)

Therefore,

\( \operatorname{Im}(u) = \ker(v). \)

Conclusion

Thus, the sequence

\( M' \xrightarrow{u} M \xrightarrow{v} M'' \rightarrow 0 \)

is exact if and only if, for every \( A \)-module \( N \), the induced sequence

\( 0 \rightarrow \operatorname{Hom}(M'',N) \rightarrow \operatorname{Hom}(M,N) \rightarrow \operatorname{Hom}(M',N) \)

is exact. This result highlights the left-exactness of the Hom functor and plays a fundamental role in homological algebra.