Hom is left exact - Knowlpedia

Hom is left exact: Let $R$ be a ring, let $M$ and $N$ be left $R$ -modules, and write $\mathrm{Hom}_R(-,-)$ for the Hom functor, with $u_*(f)=u\circ f$ and $u^*(g)=g\circ u$ .

Covariant form. If

0 \to A' \xrightarrow{u} A \xrightarrow{v} A''

is an exact sequence of left $R$ -modules, then

0 \to \mathrm{Hom}_R(M,A') \xrightarrow{u_*} \mathrm{Hom}_R(M,A) \xrightarrow{v_*} \mathrm{Hom}_R(M,A'')

is exact.

Contravariant form. If

A' \xrightarrow{u} A \xrightarrow{v} A'' \to 0

is an exact sequence of left $R$ -modules, then

0 \to \mathrm{Hom}_R(A'',N) \xrightarrow{v^*} \mathrm{Hom}_R(A,N) \xrightarrow{u^*} \mathrm{Hom}_R(A',N)

is exact.

Equivalently, $\mathrm{Hom}_R(M,-)$ preserves kernels and $\mathrm{Hom}_R(-,N)$ turns cokernels into kernels. Neither functor need preserve surjections; see the examples.

Failure of right exactness and Ext

The failure of $\mathrm{Hom}$ to be exact is measured by Ext: $\mathrm{Ext}^1_R(-,N)$ (and higher $\mathrm{Ext}^n$ ) are the right derived functors of $\mathrm{Hom}_R(-,N)$ , and every short exact sequence gives a long exact sequence in Ext (a special case of the long exact sequence for derived functors).

If $M$ is projective, then $\mathrm{Hom}_R(M,-)$ is exact. If $N$ is injective, then $\mathrm{Hom}_R(-,N)$ is exact.

Examples

In the cyclic-group examples, let $n\ge 2$ be an integer.

Example 1: Hom need not be right exact (over $\mathbb Z$ )

Start from the short exact sequence

0 \to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathbb Z/n \to 0.

Apply $\mathrm{Hom}_{\mathbb Z}(-,\mathbb Z)$ . Since $\mathrm{Hom}(\mathbb Z/n,\mathbb Z)=0$ and $\mathrm{Hom}(\mathbb Z,\mathbb Z)\cong \mathbb Z$ , we get

0 \to 0 \to \mathbb Z \xrightarrow{\times n} \mathbb Z \to \mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,\mathbb Z)\to 0.

The map $\mathbb Z\xrightarrow{\times n}\mathbb Z$ is not surjective for $|n|>1$ , so $\mathrm{Hom}(-,\mathbb Z)$ fails to preserve the cokernel; the cokernel is

\mathrm{Ext}^1_{\mathbb Z}(\mathbb Z/n,\mathbb Z)\cong \mathbb Z/n.

Example 2: If $M$ is free, then Hom is exact

If $M\cong R^{\oplus r}$ is free of finite rank $r$ , then

\mathrm{Hom}_R(M,A)\cong A^{\oplus r}.

Thus $\mathrm{Hom}_R(M,-)$ is a finite product of the identity functor, hence exact (it preserves both kernels and cokernels).

Example 3: Vector spaces over a field

If $k$ is a field and $V$ is a $k$ -vector space, then $V$ is free (hence projective). Therefore $\mathrm{Hom}_k(V,-)$ is exact, and equivalently

\mathrm{Ext}^n_k(V,W)=0 \quad (n>0)

for all $k$ -vector spaces $W$ .

Example 1: Hom need not be right exact (over Z\mathbb ZZ)

Example 2: If MMM is free, then Hom is exact

Example 3: Vector spaces over a field

Example 1: Hom need not be right exact (over $\mathbb Z$ )

Example 2: If $M$ is free, then Hom is exact