Additive functor on a long exact sequence If $$0\to X_0\to X_1\to X_2\to X_3\to \dotsb$$ is exact. Why does an additive, left-exact covariant functor $G$ gives us an exact sequence:
$$0\to G(X_0)\to G(X_1)\to G(X_2)$$
I know that by left-exactness is preserves monomorphisms so that
$$0\to G(X_0)\to G(X_1)$$
should be exact, but why exactness around $G(X_1)$?
Should I just think about changing the map out of $X_2$ to $X_2\to 0$?
 A: For an additive functor $G \colon \mathcal{A} \to \mathcal{B}$ the following two definitions of left-exactness are equivalent:


*

*For every short exact sequence $0 \to A'' \to A \to A' \to 0$ in $\mathcal{A}$, the induced sequence
$$
  0 \to G(A'') \to G(A) \to G(A')
$$
in $\mathcal{B}$ is again exact.

*For every exact sequence $0 \to A'' \to A \to A'$ in $\mathcal{A}$, the induced sequence
$$
  0 \to G(A'') \to G(A) \to G(A')
$$
in $\mathcal{B}$ is again exact.
A proof of the equivalence of the two definitions (namely the implication $1 \implies 2$) can be found here.
Note that left-exactness is (in general) much stronger than just preserving monomorphisms (see here for an example of an additive functor which preserves monomorphisms but is not left-exact).
We can solve the problem by applying the second characterization of left-exactness:
It follows from the exactness of
$$
  0 \to X_0 \to X_1 \to X_2 \to X_3 \to \dotsb
$$
that the truncated sequence
$$
  0 \to X_0 \to X_1 \to X_2
$$
is exact, from which it then follows by the left-exactness of $G$ that the sequence
$$
  0 \to G(X_0) \to G(X_1) \to G(X_2)
$$
is exact.
