# Constructing an exact sequence from a monomorphism using projective resolutions

Let $$\mathcal{A}$$ be an abelian category (hence every object in $$\mathcal{A}$$ has a projective resolution), let $$M$$ and $$N$$ be two objects in $$\mathcal{A}$$. Consider the following commutative diagram where each row is a projective resolution and the existence of $$f_0, f_1, \dots$$ is given by comparison theorem.

Do we have the following or similar conclusion?

The map $$f$$ is a monomorphism if and only if the following sequence is exact: $$\cdots \xrightarrow{} P_1\oplus Q_2 \xrightarrow{\left[ \begin{smallmatrix} \epsilon _1& 0\\ -f_1& \eta _2\\ \end{smallmatrix} \right] } P_0\oplus Q_1 \xrightarrow{\left[ \begin{smallmatrix} f_0& \eta_1 \end{smallmatrix} \right] } Q_0.$$

And how to prove that?

Enkidu told me I should consider Horseshoe lemma, but I don't know what to do.

Thank you very much.

• If you just take $M$ and all the $P_i$ to be zero, your map is a monomorphism, but the sequence isn't necessarily exact. – Achim Krause Nov 18 '20 at 6:27
• @AchimKrause It is, because it is assumed to be a projective resolution of $N$. – Matthew Pressland Nov 18 '20 at 10:19
• Well, I meant that it still has homology in degree $0$ (the original sequence is exact everywhere, including degree $0$, if and only if $M\to N$ was an isomorphism). But I now realize that the question probably asked for "exact in positive degrees". – Achim Krause Nov 18 '20 at 14:59
• Ah, ok! :) I took the lack of an extra 0 on the right to mean that exactness at the Q_0 term was not required, and then everything is consistent. If you did ask for exactness there as well then your comment is correct, of course. – Matthew Pressland Nov 19 '20 at 9:27

Note that not every abelian category has enough projectives (e.g. $$\mathscr A = \mathbf{FAb}$$, the category of finite abelian groups, does not), but of course you're free [no pun intended] to assume that $$\mathscr A$$ has enough projectives.
Note that the complex $$\cdots \to P_1 \oplus Q_2 \to P_0 \oplus Q_1 \to M \oplus Q_0 \to N \to 0$$ is exact, being the totalisation of a double complex with exact rows (the maps are the same matrices you wrote down, with the correct signs to make it a chain complex). It contains $$0 \to M \to N \to 0$$ as a subcomplex, and the quotient is the complex you describe. This gives a short exact sequence of chain complexes (written vertically): $$\begin{array}{ccccccccccc} & & 0 & & 0 & & 0 & & 0 & & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow \\ \cdots & \to & 0 & \to & 0 & \to & M & \to & N & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow \\ \cdots & \to & P_1 \oplus Q_2 & \to & P_0 \oplus Q_1 & \to & M \oplus Q_0 & \to & N & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow\\ \cdots & \to & P_1 \oplus Q_2 & \to & P_0 \oplus Q_1 & \to & Q_0 & \to & 0 & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow \\ & & 0 & & 0 & & 0 & & 0 & & 0.\!\end{array}$$ Taking the long exact homology sequence gives the result: the bottom sequence is exact at $$P_0 \oplus Q_1$$ if and only if $$M \to N$$ is injective. $$\square$$