Let us restrict to the category of modules. I'm thinking about the definition of exactness of a functor on two variables. The usual definition is that it is exact in each of the two variable, whereas I am thinking about the following definition:
For any exact sequence of pairs $$0\to (A,C)\to(A',C')\to(A'',C'')\to 0$$ we have the exact sequence $$0\to T(A,C)\to T(A',C')\to T(A'',C'')\to 0.$$
I am investigating whether this coincides with the usual definition, which mounts to solve the following problem:
If we have commutative diagram with exact rows and columns,
does it follow that $0\to A\to E\to I\to 0$ is exact? I am stuck at showing that the sequence is exact at $E$. I'm wondering whether it is true at all.
This is almost never true. For example, if $0 \to A \to B \to C \to 0$ and $0 \to A' \to B' \to C' \to 0$ are exact sequences of vector spaces, then $0 \to A \otimes A' \to B \otimes B' \to C \otimes C' \to 0$ does not have to be exact, which can be already read off the Euler characteristic: if $a=\dim(A),b=\dim(B)$ etc., why should $b=a+c$ and $b'=a'+c'$ imply $bb'=aa' + cc'$? Instead, we have that $bb'=aa'+cc' + ac'+ca'$, and that $$(A \otimes B') \oplus (B \oplus A') \to B \otimes B' \to C \otimes C' \to 0$$ is exact. In general, if $F$ is a functor in two variables with is right exact in each variable, then for exact sequences $A \to B \to C \to 0$ and $A' \to B' \to C' \to 0$ it follows that $$F(A,B') \oplus F(B,A') \to F(B,B') \to F(C,C') \to 0$$ is exact.