Editing Exact functor (section)

== Examples ==

Every [[equivalence of categories|equivalence or duality]] of abelian categories is exact.

The most basic examples of left exact functors are the [[Hom functor]]s: if '''A''' is an abelian category and ''A'' is an object of '''A''', then ''F''<sub>''A''</sub>(''X'') = Hom<sub>'''A'''</sub>(''A'',''X'') defines a covariant left-exact functor from '''A''' to the [[category of abelian groups|category '''Ab''' of abelian groups]].<ref>Jacobson (2009), p. 98, Theorem 3.1.</ref> The functor ''F''<sub>''A''</sub> is exact if and only if ''A'' is [[projective module|projective]].<ref>Jacobson (2009), p. 149, Prop. 3.9.</ref> The functor ''G''<sub>''A''</sub>(''X'') = Hom<sub>'''A'''</sub>(''X'',''A'') is a contravariant left-exact functor;<ref>Jacobson (2009), p. 99, Theorem 3.1.</ref> it is exact if and only if ''A'' is [[injective module|injective]].<ref>Jacobson (2009), p. 156.</ref>

If ''k'' is a [[field (mathematics)|field]] and ''V'' is a [[vector space]] over ''k'', we write ''V''&thinsp;* = Hom<sub>''k''</sub>(''V'',''k'') (this is commonly known as the [[dual space]]). This yields a contravariant exact functor from the [[category of vector spaces|category of ''k''-vector spaces]] to itself. (Exactness follows from the above: ''k'' is an [[injective module|injective]] ''k''-[[module (mathematics)|module]]. Alternatively, one can argue that every short exact sequence of ''k''-vector spaces [[split exact sequence|splits]], and any additive functor turns split sequences into split sequences.)

If ''X'' is a [[topological space]], we can consider the abelian category of all [[sheaf (mathematics)|sheaves]] of [[abelian group]]s on ''X''. The covariant functor that associates to each sheaf ''F'' the group of global sections ''F''(''X'') is left-exact.

If ''R'' is a [[ring (mathematics)|ring]] and ''T'' is a right ''R''-[[module (mathematics)|module]], we can define a functor ''H''<sub>''T''</sub> from the abelian [[Category of modules|category of all left ''R''-modules]] to '''Ab''' by using the [[tensor product]] over ''R'': ''H''<sub>''T''</sub>(''X'') = ''T'' ⊗ ''X''. This is a covariant right exact functor; in other words, given an exact sequence ''A''→''B''→''C''→0 of left ''R'' modules, the sequence of abelian groups ''T'' ⊗ ''A'' → ''T'' ⊗ ''B'' → ''T'' ⊗ ''C'' → 0 is exact. 

The functor ''H''<sub>''T''</sub> is exact if and only if ''T'' is [[flat module|flat]]. For example, <math>\mathbb{Q}</math> is a flat <math>\mathbb{Z}</math>-module. Therefore, tensoring with <math>\mathbb{Q}</math> as a <math>\mathbb{Z}</math>-module is an exact functor. '''Proof:''' It suffices to show that if ''i'' is an [[injective map]] of <math>\mathbb{Z}</math>-modules <math>i:M\to N</math>, then the corresponding map between the tensor products <math>M \otimes \mathbb{Q} \to N\otimes \mathbb{Q}</math> is injective. One can show that <math>m \otimes q = 0</math> if and only if <math>m</math> is a torsion element or <math>q = 0</math>. The given tensor products only have pure tensors. Therefore, it suffices to show that if a pure tensor <math>m \otimes q </math> is in the [[kernel (algebra)|kernel]], then it is zero. Suppose that <math>m \otimes q </math> is an element of the kernel. Then, <math>i(m)</math> is torsion. Since <math>i</math> is injective, <math>m</math> is torsion. Therefore, <math>m \otimes q  = 0</math>. Therefore, <math> M \otimes \mathbb{Q} \to N\otimes \mathbb{Q} </math> is also injective.

In general, if ''T'' is not flat, then tensor product is not left exact. For example, consider the short exact sequence of <math>\mathbf{Z}</math>-modules <math>5\mathbf{Z} \hookrightarrow \mathbf{Z} \twoheadrightarrow \mathbf{Z}/5\mathbf{Z}</math>. Tensoring over <math>\mathbf{Z}</math> with <math>\mathbf{Z}/5\mathbf{Z}</math> gives a sequence that is no longer exact, since <math>\mathbf{Z}/5\mathbf{Z}</math> is not torsion-free and thus not flat. 

If '''A''' is an abelian category and '''C''' is an arbitrary [[small category|small]] [[category (mathematics)|category]], we can consider the [[functor category]] '''A<sup>C</sup>''' consisting of all functors from '''C''' to '''A'''; it is abelian. If ''X'' is a given object of '''C''', then we get a functor ''E''<sub>''X''</sub> from '''A'''<sup>'''C'''</sup> to '''A''' by evaluating functors at ''X''. This functor ''E''<sub>''X''</sub> is exact.

While tensoring may not be left exact, it can be shown that tensoring is a right exact functor: 

Theorem: Let ''A'',''B'',''C'' and ''P'' be ''R''-modules for a [[commutative ring]] ''R'' having multiplicative identity. Let <math>A \ \stackrel{f}{\to} \ B\ \stackrel{g}{\to} \ C \to 0</math> be a [[short exact sequence]] of  ''R''-modules. Then 
:<math>A\otimes_{R} P \stackrel{f \otimes P}\to B\otimes_{R} P \stackrel{g \otimes P}\to C \otimes_{R} P \to 0</math>
is also a short exact sequence of ''R''-modules. (Since ''R'' is commutative, this sequence is a sequence of ''R''-modules and not merely of abelian groups). Here, we define
:<math>f \otimes P(a \otimes p):=f(a) \otimes p, g \otimes P(b \otimes p):=g(b) \otimes p</math>. 

This has a useful [[corollary]]: If ''I'' is an [[ideal (ring theory)|ideal]] of ''R'' and ''P'' is as above, then <math>P \otimes_{R} (R/I) \cong P/IP</math>.

Proof: <math> I \stackrel{f}\to  R \stackrel{g}\to R/I \to 0</math>, where ''f'' is the inclusion and ''g'' is the projection, is an exact sequence of ''R''-modules. By the above we get that :<math>I\otimes_{R} P \stackrel{f \otimes P}\to R\otimes_{R} P \stackrel{g \otimes P}\to R/I \otimes_{R} P \to 0</math> is also a short exact sequence of ''R''-modules. By exactness, <math>R/I \otimes_{R} P \cong (R\otimes_{R} P)/Image(f\otimes P) = (R\otimes_{R} P)/(I \otimes_{R} P)</math>, since ''f'' is the inclusion. Now, consider the [[module homomorphism|''R''-module homomorphism]] from <math>R \otimes_R P \rightarrow P</math> given by ''R''-linearly extending the map defined on pure tensors: <math>r\otimes p \mapsto rp. rp=0 </math> implies that <math>0= rp\otimes 1 = r \otimes p</math>. So, the kernel of this map cannot contain any nonzero pure tensors.  <math>R \otimes_R P</math> is composed only of pure tensors: For  <math> x_i \in R,  \sum_{i} x_i (r_i \otimes p_i) = \sum_i 1 \otimes (r_i x_i p_i) = 1 \otimes (\sum_i  r_i x_i p_i)</math>. So, this map is injective. It is clearly [[surjective|onto]]. So, <math>R \otimes_R P \cong P</math>. Similarly, <math>I \otimes_R P \cong IP</math>. This proves the corollary.

As another application, we show that for, <math>P =\mathbf{Z}[1/2]:= \{a/2^k : a,k \in \mathbf{Z}\}, P \otimes \mathbf{Z}/m\mathbf{Z} \cong P/k\mathbf{Z}P </math> where <math> k=m/2^n </math> and ''n'' is the highest [[power of 2]] dividing ''m''. We prove a special case: ''m''=12.

Proof: Consider a pure tensor <math>(12z)\otimes (a/2^k ) \in  
 (12\mathbf{Z} \otimes_{Z} P).(12z)\otimes (a/2^k ) = (3z)\otimes (a/2^{k-2}) </math>. Also, for <math>(3z)\otimes (a/2^k ) \in  
 (3\mathbf{Z} \otimes_{Z} P), (3z)\otimes (a/2^k ) = (12z)\otimes (a/2^{k+2}) </math>.
This shows that <math>(12\mathbf{Z} \otimes_{Z} P) = (3\mathbf{Z} \otimes_{Z} P)</math>. Letting <math>P= \mathbf{Z}[1/2], A = 12\mathbf{Z}, B= \mathbf{Z}, C = \mathbf{Z}/12\mathbf{Z} </math>,  ''A,B,C,P'' are ''R''='''Z''' modules by the usual multiplication action and satisfy the conditions of the main [[theorem]]. By the exactness implied by the theorem and by the above note we obtain that
<math>: \mathbf{Z}/12\mathbf{Z} \otimes_{Z} P \cong (\mathbf{Z} \otimes_{Z} P) / (12\mathbf{Z} \otimes_{Z} P) = (\mathbf{Z} \otimes_{Z} P) / (3\mathbf{Z} \otimes_{Z} P) \cong \mathbf{Z}P/3\mathbf{Z} P </math>. The last congruence follows by a similar argument to one in the proof of the corollary showing that <math> I \otimes_R P \cong IP </math>.