Vtyjkyuk

If $$0to X_0to X_1to X_2to X_3to dotsb$$ is exact. Why does an additive, left-exact covariant functor $G$ gives us an exact sequence:
$$0to G(X_0)to G(X_1)to G(X_2)$$
I know that by left-exactness is preserves monomorphisms so that
$$0to G(X_0)to G(X_1)$$
should be exact, but why exactness around $G(X_1)$?

Should I just think about changing the map out of $X_2$ to $X_2to 0$?

For an additive functor $G colon mathcalA to mathcalB$ the following two definitions of left-exactness are equivalent:

1. For every short exact sequence $0 to A'' to A to A' to 0$ in $mathcalA$, the induced sequence
   $$
   0 to G(A'') to G(A) to G(A')
   $$
   in $mathcalB$ is again exact.




2. For every exact sequence $0 to A'' to A to A'$ in $mathcalA$, the induced sequence
   $$
   0 to G(A'') to G(A) to G(A')
   $$
   in $mathcalB$ 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.