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Cohomologies.

Yoshifumi Tsuchimoto

\fbox{02. \lq\lq Hom'' modules.}

LEMMA 02.1   Let $ R$ be a ring. Let $ f:M\to N$ be a homomorphism of $ R$ -modules. Then for any $ R$ -module $ L$ we may define:
  1. A homomorphism $ \operatorname{Hom}_R(L,f): \operatorname{Hom}_R(L,M)\to \operatorname{Hom}_R(L,N)$ defined by $ \operatorname{Hom}_R(L,f)(g)=f\circ g$ .
  2. A homomorphism $ \operatorname{Hom}_R(f,L): \operatorname{Hom}_R(N,L)\to \operatorname{Hom}_R(M,L)$ defined by $ \operatorname{Hom}_R(f,L)(h)=h\circ f$ .

PROPOSITION 02.2   Let $ R$ be a ring. Let

$\displaystyle 0 \to M_1 \overset{f}{\to} M_2 \overset{g}{\to} M_3 \to 0
$

be an exact sequence of $ R$ -modules. Then for any $ R$ -module $ N$ , we have:
  1. $\displaystyle 0 \to \operatorname{Hom}_R (N,M_1)
\overset{\operatorname{Hom}_R...
...(N,M_2)
\overset{\operatorname{Hom}_R(N,g)}{\to} \operatorname{Hom}_R(N,M_3)
$

    is exact. The third arrow $ \operatorname{Hom}_R(N,g)$ need not be surjective.
  2. $\displaystyle 0 \to \operatorname{Hom}_R (M_3,N)
\overset{\operatorname{Hom}_R...
...(M_2,N)
\overset{\operatorname{Hom}_R(f,N)}{\to} \operatorname{Hom}_R(M_1,N)
$

    is exact. The third arrow $ \operatorname{Hom}_R(f,N)$ need not be surjective.

EXERCISE 02.1   We consider an exact sequence

$\displaystyle 0\to 3\mathbb{Z}\overset{i}{\to}\mathbb{Z}\to \mathbb{Z}/3 \mathbb{Z}\to 0
$

where $ i$ is the inclusion map. Show that

$\displaystyle \operatorname{Hom}_\mathbb{Z}(\mathbb{Z},\mathbb{Z}) \overset{\op...
...Hom}(i,\mathbb{Z})}{\to} \operatorname{Hom}_\mathbb{Z}(3\mathbb{Z},\mathbb{Z})
$

is not surjective

EXERCISE 02.2   Assume $ R$ is a field. Then show that the third arrow which appear in the sequence (1) in Proposition 2.2 is surjective.


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2010-04-15