pullback is preserved by left exact covariant functor $T : \mathsf{Mod}_R \to \mathsf{Ab}$

This is a problem in Rotman Homological Algebra, 5.16. The following is the statement of the problem.

Prove every left exact covariant functor $T:\mathsf{Mod}_R \to \mathsf{Ab}$ preserves pullbacks. Especially $\operatorname{Hom}(M,-)$ preserves pullback.

$\operatorname{Hom}(M,-)$ preserves pull back can be proven very easily by additive property of $\operatorname{Hom}(M,B\oplus C)=\operatorname{Hom}(M,B)\oplus \operatorname{Hom}(M,C)$. Left exactness of $\operatorname{Hom}(M,-)$ enforces commutativity of the pull back diagram.(It even suffices to assume $B,C\subset A$ and pullback of $A$ along $B,C$ to prove the problem as only the some submodules of $A,B,C$ matters for commutativity of pullback.)

However the statement of the problem did not state $T$ is additive at all. It might be the case that the pullback of $T(A)$ along $T(B),T(C)$ might not be a subgroup of $T(B)\oplus T(C)$ where $A,B,C\in \mathsf{Mod}_R$. There is no reason to believe $T(A\oplus B)=T(A)\oplus T(B)$ here.

How should I prove the general case without the assumption of additive property?

• A left exact functor commutes with finite limits, in particular with finite products, and thus is always additive. – Roland Jun 12 '17 at 7:36

We need the following: $$\require{AMScd} \begin{CD} A @>>> B \\ @VVV @VVV \\ C @>>> D \end{CD}$$

This is pullback iff $0\rightarrow A \rightarrow B\oplus C \rightarrow D$ is exact.

• To future readers: be careful. While this fact is important to do the exercise, it still requires additivity to use this to complete the proof. Luckily, as Roland points out in the comments, left exact functors are always additive. See this related question I just asked, which Roland answered in the comments for more details. – jgon Jan 7 at 22:06

Let $0\to A_1\to A_1\oplus A_2\to A_2\to 0$ be exact and $T:Ab\to Ab$ be left exact.

$0\to TA_1\to T(A_1\oplus A_2)\to TA_2$. The point is to show at $T(A_1\oplus A_2)\to TA_2$ is surjective. Since $A_2$ is a direct summand of $A_1\oplus A_2$, I got a retraction $\pi_2:A_1\oplus A_2\to A_2,i_2:A_2\to A_1\oplus A_2$ where $i_2$ is just injection and $\pi_2$ is just projection. So $T(1_{A_2})=1_{T(A_2)}=T(\pi_2\circ i_2)=T\pi_2\circ Ti_2$. Since $Ti_2$ is injection from T being left exact, I have found a preimage of any $Ti_2(a_2)\in A_1\oplus A_2$. Hence it is surjective.

So for the right exact case, the argument is similar and it suffices to check retraction giving $0\in A_1$. This also shows right exact preserving finite limit.

• @Roland, would you mind check my answer for its correctness? Thanks. – user45765 Jun 13 '17 at 22:23
• @Sky would you mind check my answer for its correctness? Thanks. – user45765 Jun 13 '17 at 22:23
• it is no need to check this. You can do this:for any split exact sequence and additive functor $T$, $T$ preserves split exact sequence. $T$ need not be left or right exact. – Sky Jun 14 '17 at 8:38
• @Sky Yes. If it is additive, I do not need left or right exact at all and the split is very natural. – user45765 Jun 14 '17 at 14:47