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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.

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    $\begingroup$ 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. $\endgroup$ – Achim Krause Nov 18 '20 at 6:27
  • $\begingroup$ @AchimKrause It is, because it is assumed to be a projective resolution of $N$. $\endgroup$ – Matthew Pressland Nov 18 '20 at 10:19
  • $\begingroup$ 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". $\endgroup$ – Achim Krause Nov 18 '20 at 14:59
  • $\begingroup$ 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. $\endgroup$ – Matthew Pressland Nov 19 '20 at 9:27
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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. $\square$

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  • $\begingroup$ Thank you very much! $\endgroup$ – gcfsjfcus Nov 18 '20 at 7:02

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