If $$ 0 \to A \to B\to C,$$ is a left exact sequence of $R$-module, then for any $R$-module $M$, $$ 0 \to \operatorname{Hom}_R(M,A)\to \operatorname{Hom}_R(M,B)\to \operatorname{Hom}_R(M,C), $$ is left exact.

I proved the above, and highlighted what I'm a little unfamiliar with: Let $$ 0 \to A\ \xrightarrow{i}\ B\ \xrightarrow{f}\ C, $$ and $$ 0 \to \operatorname{Hom}(M,A)\ \xrightarrow{\operatorname{Hom}(M,i)}\ \operatorname{Hom}(M,B)\ \xrightarrow{\operatorname{Hom}(M,f)}\ \operatorname{Hom}(M,C). $$ We need to show that $$ \ker\left(\operatorname{Hom}(M,f)\right)=\operatorname{Hom}(M,i)(\operatorname{Hom}(M,A)). $$

Let $i \circ \varphi \in \operatorname{RHS}$. Then $f \circ i \circ \varphi : M \to C$ is $0$ since $f \circ i \circ \varphi(M) \subseteq f( i(A)) = f(\ker(f))=0$.

Conversely, let $\psi \in \operatorname{LHS}$. Then $f \circ \psi = 0$ so that $ f(\psi(M))=0$. Hence $\psi(M) \subseteq \ker(f)=i(A)$. Since the image of $\psi$ is contained in the image of $i$, we may factor $\psi$ as $\psi=i \varphi$ with $\varphi : M \to A$. (Here is my trial, but I'm not fully understanding this: Since $i$ is injective, $i(A)$ is isomorphic with $A$. So $i^{-1}(\psi (M)) \subseteq A$ and if we let $\varphi=i^{-1} \psi$, then $\psi = i \varphi$.)

And I have one more question: The above looks very messy, especially the notation. Is there a better proof/understanding about it?

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    $\begingroup$ Re: notation. Why not use $i_{\ast}$ and $f_{\ast}$ instead of $Hom_{R}(M,i)$ and $Hom_{R}(M,f)$? $\endgroup$ – t.b. Jun 24 '11 at 13:23
  • $\begingroup$ This is supposed to follow from the fact that $\text{Hom}(-, A)$ preserves limits, but I am not sure about the details. $\endgroup$ – Qiaochu Yuan Jun 24 '11 at 13:24
  • $\begingroup$ If $N$ is an $A$-module, then we can consider the functors (from the category of $A$-modules to itself) $M\to M\otimes N$ and $P\to \text{Hom}(N,P)$ which we denote by $T$ and $U$, respectively. We then have $\text{Hom}(T(M),P)=\text{Hom}(M,U(P))$. In particular, we say that $U$ is a $\textbf{right adjoint}$ of $T$. It is a general fact that any functor which is a right adjoint is left exact. Therefore, $U$ is a left exact functor. $\endgroup$ – Amitesh Datta Jun 24 '11 at 14:02
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    $\begingroup$ @Amitesh: yes, that's indeed true. In fact, you can get the left exactness from abstract nonsense simply by recalling what the definition of a limit means and saying $\operatorname{Hom}(M,-)$ preserves limits tautologically. $\endgroup$ – t.b. Jun 24 '11 at 14:38
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    $\begingroup$ Dear Gobi, Theo's explanation below is (as usual) very helpful. I think it is also worth emphasizing that this is very important: in fact, it is essentially the definition of exactness of such a sequence in an abelian category! Or, restated, it is the universal property of the kernel: a map $X \to B$ factors through $A$ if and only if the composite $X \to B \to C$ is zero (and then the factorization through $A$ is unique). $\endgroup$ – Akhil Mathew Jun 24 '11 at 21:36

Well, your proof is okay. Let me suggest a slightly different way of looking at it:

Consider a sequence

$$ 0\; \xrightarrow{\phantom{ij}} \; A \; \xrightarrow{i\phantom{j}}\; B\; \xrightarrow{j\phantom{i}} \; C$$ and look at $$ 0\; \xrightarrow{\phantom{j_{\ast}}} \operatorname{Hom}{(M,A)}\; \xrightarrow{i_{\ast}}\operatorname{Hom}{(M,B)} \;\xrightarrow{j_{\ast}}\; \operatorname{Hom}{(M,C)},$$ where I write $i_{\ast} = \operatorname{Hom}(M,i)$ and $j_{\ast} = \operatorname{Hom}(M,j)$.

Saying that the first sequence is exact amounts to saying $i = \ker{j}$, that is $ji = 0$ and $i$ has the universal property as depicted in the first diagram below: If $g: M \to B$ is such that $jg = 0$ then there exists a unique $f: M \to A$ such that $if = g$. In other words if $j_\ast g = 0$ then $g = i_{\ast}f$, or yet again $\operatorname{Ker}{j_\ast} \subset \operatorname{Im}{i_{\ast}}$ and $i_{\ast}$ is injective.

property kernel/cokernel

On the other hand, the second diagram says: if $g: M \to B$ is of the form $g = if = i_{\ast}f$ then $j_{\ast}g = 0$ (because $j_\ast g = j_{\ast}i_{\ast} f = (ji)_{\ast}f = 0f = 0$). In other words, $\operatorname{Im}{i_{\ast}} \subset \operatorname{Ker}{j_{\ast}}$.

Summing up, we have shown that for all $M$ the sequence

$$ 0\; \xrightarrow{\phantom{j_{\ast}}} \operatorname{Hom}{(M,A)}\; \xrightarrow{i_{\ast}}\operatorname{Hom}{(M,B)} \;\xrightarrow{j_{\ast}}\; \operatorname{Hom}{(M,C)}$$ is exact both at $\operatorname{Hom}{(M,A)}$ ($i_{\ast}$ is injective) and at $\operatorname{Hom}{(M,B)}$ ($\operatorname{Im}{i_{\ast}} = \operatorname{Ker}{j_{\ast}}$)—you seem to have forgotten about the first point here.

Added: As witnessed by the argument above, left exactness of $\operatorname{Hom}$ is essentially the definition of left exactness in the abelian category of $R$-modules. As the comments try to point out, the importance of this fact cannot be overemphasized.

I would like to add two further points:

  1. A functor $F$ is left exact in your definition if and only if $0 \to F(A) \to F(B) \to F(C)$ is left exact for every short exact sequence $0 \to A \to B \to C \to 0$.

    Indeed, in a left exact sequence $0 \to A \to B \to C$, we may factor $j: B \to C$ over its image as $B \twoheadrightarrow \operatorname{Im}{j} \rightarrowtail C$ and obtain two exact sequences $$0 \to A \to B \to \operatorname{Im}{j} \to 0 \qquad \text{and} \qquad 0 \to \operatorname{Im}{j} \to C \to \operatorname{Coker}{j} \to 0.$$ Applying $F$ to these two exact sequences, we obtain the left exact sequences $$0 \to F(A) \to F(B) \to F(\operatorname{Im}{j}) \qquad \text{and} \qquad 0 \to F(\operatorname{Im}{j}) \to F(C ) \to F(\operatorname{Coker}{j}).$$ Since the kernel of a map is not changed by postcomposing the map with a monomorphism (check this!), we have $$\operatorname{Ker}{(F(B) \to F(\operatorname{Im}{j}))} = \operatorname{Ker}{(F(B) \to F(\operatorname{Im}{j}) \to F(C))},$$ so by functoriality of $F$ we get a left exact sequence $0 \to F(A) \to F(B) \to F(C)$ as desired.

  2. A natural question is: When does $\operatorname{Hom}(M,-)$ send short exact sequences to short exact sequences? In other words, when is $j_\ast = \operatorname{Hom}{(M,j)}$ an epimorphism for all short exact sequences $0\; \xrightarrow{\phantom{ij}} \; A \; \xrightarrow{i\phantom{j}}\; B\; \xrightarrow{j\phantom{i}} \; C \to 0$?

    In view of left exactness of $\operatorname{Hom}{(M,-)}$ the question is: Given any morphism $h: M \to C$ and any epimorphism $j: B \to C$, when is $h$ of the form $h = j_\ast g$ for some morphism $g: M \to B$?
    Lifting property
    As you certainly know, this is precisely the definition of projective modules: $M$ is called projective if and only if $g$ always exists, for all epimorphisms $j: B \twoheadrightarrow C$ and all $h: M \to C$. For emphasis:

    A module $M$ is projective if and only if $\operatorname{Hom}{(M,-)}$ is exact, that is: it sends short exact sequences to short exact sequences.

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    $\begingroup$ @Amitesh: much appreciated, thanks! $\endgroup$ – t.b. Jun 24 '11 at 14:38
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    $\begingroup$ Thanks for your great answer! $\endgroup$ – Gobi Jun 25 '11 at 8:48
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    $\begingroup$ @Gobi: You're welcome. Glad I could help! $\endgroup$ – t.b. Jun 25 '11 at 12:15
  • $\begingroup$ The half-exactness of Hom is the essential "practical" ingredient in proof by a ("small") Yoneda's lemma of the (suitable) half-exactness of left/right adjoint functors. $\endgroup$ – paul garrett Jun 26 '11 at 23:04
  • $\begingroup$ How are we showing left-exact in $(1)$ without showing $\ker \subset Im$ and $Im \subset \ker$? $\endgroup$ – Al Jebr Oct 20 '19 at 7:11

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