Say we have a ring $R$ and let $A,B,C$ be $R$-mod. Then prove $\hom_R(N,-)$ is left exact (where $N$ is some fixed $R$-mod).

Basically we want to show that given that we know that $0\rightarrow A\stackrel{f}{\longrightarrow}B\stackrel{g}{\longrightarrow} C$ is exact, we want to show that their images under this functor satisfy the same property. That is, we want to show that $$0\rightarrow \hom_R(N,A)\stackrel{\sigma}{\longrightarrow} \hom_R(N,B)\stackrel{\tau}{\longrightarrow} \hom_R(N,C)$$is exact.

First of all I assume that the functor $\hom_R(N,-)$ takes the arrow $f$ in the category of $R$-modules to $\sigma$ (an arrow in the category of abelian groups), where for $\phi\in\hom_R(N,A)$, we have $\sigma(\phi)=f\circ\phi$. Similarly we define $\tau(\gamma)=g(\gamma)$ where $\gamma\in \hom_R(N,B)$. Then to prove that this is sequence is exact we have to prove two things. $\sigma$ is injective, and second $\text{im}(\sigma)=\ker(\tau)$.

The first part is easy since we have that if $\sigma(\phi)=0$, then for all $n$, we have that $f(\phi(n))=0$, but as $f$ is injective, we have that for all $n$, $\phi(n)=0$, so this means $\phi=0$, and thus $\sigma$ is injective.

Now let $\gamma\in \text{im}(\sigma)$. Then, there exists a $\phi\in \hom_R(N,A)$ such that $\sigma(\phi)=\gamma$. Then $\tau(\gamma)=g(\gamma)=g(\sigma(\phi))=g(f(\phi))=0$, as $g\circ f$ is $0$. To prove the other inclusion, let $\gamma\in \ker(\tau)$. Then $\tau(\gamma)=0$, which means $g(\gamma)=0$. Thus, for each $n\in N$ we have that $g(\gamma(n))=0$, so $\gamma(n)\in\ker(g)=\text{im}(f)$. Hence, for each $n\in N$ there exist an $a_n\in A$ such that $\gamma(n)=f(a_n)$. I was thinking about maybe defining $\phi(n)=a_n$, if we manage to show that $\phi\in \hom_R(N,A)$ then we would be done because $\gamma(n)=f(a_n)=f(\phi(n))=\sigma(\phi(n))$, so $\gamma=\sigma(\phi)$, but I am not sure at how to go on proving that $\phi\in \hom_R(N,A)$ since I am picking $a_n$ to be an arbitrary element element in $A$ that does $f(a_n)=\gamma(n)$.

Also, after I am done hammering that small detail, how would one go at proving that this functor is not right exact?


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    $\begingroup$ It is right exact sometimes (precisely when $N$ is projective). A cleaner but less concrete way to show that this functor is left exact is to show that it has a left adjoint. $\endgroup$ – Qiaochu Yuan Dec 8 '12 at 8:51
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    $\begingroup$ I just picked Jacobson (page 149 vol II) and saw that $N$ is projective if and only the above functor is exact. $\endgroup$ – Daniel Montealegre Dec 8 '12 at 9:21

Ok I misread that you were having some trouble proving exactness at $\textrm{Hom}_R(N,B)$ firstly we note that we get for free that $\textrm{im} \sigma \subset \ker \tau$. This is because $\tau \sigma (\phi)$ for any $\phi\in \textrm{Hom}_R(N,A)$ is just $g \circ f \circ \phi $ that is zero by exactness at $B$. For the other direction suppose $\tau(\gamma) = 0$. Then $g \circ \gamma = 0$ and so the image of $\gamma$ is contained in the image of $f$. That is to say for every $a \in N$ there exists an $a' \in A$ such that $\gamma (a) = f(a')$. Now define a function $l : N \to A$ by $l(a) = a'$. This is well defined for if there is $a''$ such that $f(a') = f(a'')$ then injectivity of $f$ implies that $a'= a''$ so there is no problem. It will now follow that $\gamma = \sigma(l)$ proving exactness at $\textrm{Hom}_R(N,B)$.

I assume from your question that you want an example of an exact sequence and an $R$ - module $N$ such that applying the covariant functor $\textrm{Hom}_R(N,-)$ produces a sequence that is not exact on the right side. For a simple example we consider

$$ 0 \to \Bbb{Z} \to \Bbb{Z} \to \Bbb{Z}/n\Bbb{Z} \to 0$$

and take $N = \Bbb{Z}/n\Bbb{Z}$. Then the sequence

$$ 0 \to \textrm{Hom}_\Bbb{Z}(\Bbb{Z}/n\Bbb{Z},\Bbb{Z}) \to \textrm{Hom}_\Bbb{Z}(\Bbb{Z}/n\Bbb{Z},\Bbb{Z}) \to \textrm{Hom}_\Bbb{Z}(\Bbb{Z}/n\Bbb{Z},\Bbb{Z}/n\Bbb{Z})\to 0$$

cannot possibly be exact on the right because then we would have $\textrm{Hom}_\Bbb{Z}(\Bbb{Z}/n\Bbb{Z},\Bbb{Z}/n\Bbb{Z}) \cong 0$ which is ridiculous.

Let me now give an example for when this functor is exact. If $N$ is free of finite rank then we can write $N = \bigoplus R^i$, the direct sum of finitely many copies of $R$. Then recalling that

  1. $\textrm{Hom}$ commutes with finite direct sums and

  2. $\textrm{Hom}_R(R,M) \cong M$ for any $R$ - module $M$

will now show that $\textrm{Hom}_R(N,-)$ is an exact functor.

Edit: Now that we talk about the left exact functor hom I would just like to mention something about what we "add at the end". In theory, it would be nice if from the left exact sequence of Hom modules that we can somehow "continue" to get a long exact sequence. This is where the following result comes in:

Proposition (First LES for Ext): Given any $R$ - module $N$, every ses of $R$ - modules $0 \to M''\to M \to M'\to 0$ gives rise to a long exact sequence in Ext groups:

$$ 0 \to \textrm{Hom}_R(N,M'') \to \textrm{Hom}_R(N,M) \to \textrm{Hom}_R(N,M') \to \textrm{Ext}^1_R(N,M'') \to \textrm{Ext}^1_R(N,M) \to \textrm{Ext}^1_R(N,M') \to \ldots $$

The Ext functor is kinda complicated to define for we need to talk about projective resolutions and taking homologies of chain complexes but for now the way I think of it is as a functor from $R$ - mod to itself "derived" from the Hom functor. I should also add that the Ext functor is also important because it sort of "measures" how far away is cohomology different from taking Hom of the homology. This is the statement of the Universal Coefficients Theorem and is very important in Algebraic Topology! In fact Universal Coefficients tells you that because the ses involved is split knowing Ext and homology will completely determine the cohomology upto isomorphism.

Lastly, if you want to see how these sequences are useful you can view my post here. If you want to compute Ext with your bare hands you can view another post here.

  • $\begingroup$ This is probably going to be really dumb, but how do you get Hom$_\mathbb{Z}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z}/n\mathbb{Z})\cong 0$? $\endgroup$ – Daniel Montealegre Dec 9 '12 at 3:43
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    $\begingroup$ @DanielMontealegre There is no group homomorphism from a finite abelian group into a free abelian group because the latter is torsion free. It follows that your exact sequence looks like $0 \to 0 \to 0 \to \textrm{Hom}_\Bbb{Z}(\Bbb{Z}/n,\Bbb/n) \to 0$. Exactness at each of the three middle terms of the sequence imply that $\textrm{Hom}_\Bbb{Z}(\Bbb{Z}/n,\Bbb/n) \cong 0$. $\endgroup$ – user38268 Dec 9 '12 at 11:56

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