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

Question

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?

Answer 1

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.

Answer 2

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.