Very basic derived functor question

Question

This my first ever exposure to the notion of having "enough injectives" and to derived functors in general. I am looking at this Wikipedia page, speifically the 'Construction and first properties' section. For simplicity just assume we're in an Abelian category and the class of morphisms is all the monomorphisms. Let $X$ be an object in the category $\mathcal{C}$. How do we know that there must exist an exact sequence, $$ 0 \longrightarrow X \longrightarrow I^{0} \longrightarrow I^{1} \longrightarrow I^{2} \longrightarrow \cdots ,$$ where the $I^{i}$ objectives are injective objects? I can see how knowing that every object has a monomorphism from it to some injective object would give you exactness at the first term (trivially), but then how do you get exactness for further terms?

Answer 1

The Wikipedia page you refer to does not seem to give the whole construction, which is the following.

One starts with a monomorphism $\iota\colon X\rightarrowtail I^0$ to an injective object. Then, to obtain an exact sequence, we want to factor through the cokernel of $\iota$, so we take the canonical epimorphism $\epsilon^0\colon I^0 \twoheadrightarrow \operatorname{coker} \iota$ and compose it with a monomorphism $\iota^1\colon \operatorname{coker} \iota \rightarrowtail I^1$, which exists because we have enough injectives.

Inductively, the differential $d^n\colon I^n \to I^{n+1}$ is the composition $\iota^{n+1}\circ\epsilon^n$ of the canonical epimorphism $\epsilon^n\colon I^n \twoheadrightarrow \operatorname{coker} \iota^n$ and a monomorphism $\iota^{n+1}\colon \operatorname{coker} \iota^n\rightarrowtail I^{n+1}$.

You should be able to prove yourself that the resulting sequence is exact (e.g. by a diagram chase).