Theorem 10.82.3 (058K)—The Stacks project

$0 \to M_1 \xrightarrow {f_1} M_2 \xrightarrow {f_2} M_3 \to 0$

be an exact sequence of $R$ -modules. The following are equivalent:

The sequence $0 \to M_1 \to M_2 \to M_3 \to 0$ is universally exact.
For every finitely presented $R$ -module $Q$ , the sequence

$0 \to M_1 \otimes _ R Q \to M_2 \otimes _ R Q \to M_3 \otimes _ R Q \to 0$

is exact.
Given elements $x_ i \in M_1$ $(i = 1, \ldots , n)$ , $y_ j \in M_2$ $(j = 1, \ldots , m)$ , and $a_{ij} \in R$ $(i = 1, \ldots , n, j = 1, \ldots , m)$ such that for all $i$

$f_1(x_ i) = \sum \nolimits _ j a_{ij} y_ j,$

there exists $z_ j \in M_1$ $(j =1, \ldots , m)$ such that for all $i$ ,

$x_ i = \sum \nolimits _ j a_{ij} z_ j .$
Given a commutative diagram of $R$ -module maps

$\xymatrix{ R^ n \ar[r] \ar[d] & R^ m \ar[d] \\ M_1 \ar[r]^{f_1} & M_2 }$

where $m$ and $n$ are integers, there exists a map $R^ m \to M_1$ making the top triangle commute.
For every finitely presented $R$ -module $P$ , the $R$ -module map $\mathop{\mathrm{Hom}}\nolimits _ R(P, M_2) \to \mathop{\mathrm{Hom}}\nolimits _ R(P, M_3)$ is surjective.
The sequence $0 \to M_1 \to M_2 \to M_3 \to 0$ is the colimit of a directed system of split exact sequences of the form

$0 \to M_{1} \to M_{2, i} \to M_{3, i} \to 0$

where the $M_{3, i}$ are finitely presented.

Proof. Obviously (1) implies (2).

Next we show (2) implies (3). Let $f_1(x_ i) = \sum _ j a_{ij} y_ j$ be relations as in (3). Let $(d_ j)$ be a basis for $R^ m$ , $(e_ i)$ a basis for $R^ n$ , and $R^ m \to R^ n$ the map given by $d_ j \mapsto \sum _ i a_{ij} e_ i$ . Let $Q$ be the cokernel of $R^ m \to R^ n$ . Then tensoring $R^ m \to R^ n \to Q \to 0$ by the map $f_1: M_1 \to M_2$ , we get a commutative diagram

$\xymatrix{ M_1^{\oplus m} \ar[r] \ar[d] & M_1^{\oplus n} \ar[r] \ar[d] & M_1 \otimes _ R Q \ar[r] \ar[d] & 0 \\ M_2^{\oplus m} \ar[r] & M_2^{\oplus n} \ar[r] & M_2 \otimes _ R Q \ar[r] & 0 }$

where $M_1^{\oplus m} \to M_1^{\oplus n}$ is given by

$(z_1, \ldots , z_ m) \mapsto (\sum \nolimits _ j a_{1j} z_ j, \ldots , \sum \nolimits _ j a_{nj} z_ j),$

and $M_2^{\oplus m} \to M_2^{\oplus n}$ is given similarly. We want to show $x = (x_1, \ldots , x_ n) \in M_1^{\oplus n}$ is in the image of $M_1^{\oplus m} \to M_1^{\oplus n}$ . By (2) the map $M_1 \otimes Q \to M_2 \otimes Q$ is injective, hence by exactness of the top row it is enough to show $x$ maps to $0$ in $M_2 \otimes Q$ , and so by exactness of the bottom row it is enough to show the image of $x$ in $M_2^{\oplus n}$ is in the image of $M_2^{\oplus m} \to M_2^{\oplus n}$ . This is true by assumption.

Condition (4) is just a translation of (3) into diagram form.

Next we show (4) implies (5). Let $\varphi : P \to M_3$ be a map from a finitely presented $R$ -module $P$ . We must show that $\varphi$ lifts to a map $P \to M_2$ . Choose a presentation of $P$ ,

$R^ n \xrightarrow {g_1} R^ m \xrightarrow {g_2} P \to 0.$

Using freeness of $R^ n$ and $R^ m$ , we can construct $h_2: R^ m \to M_2$ and then $h_1: R^ n \to M_1$ such that the following diagram commutes

$\xymatrix{ & R^ n \ar[r]^{g_1} \ar[d]^{h_1} & R^ m \ar[r]^{g_2} \ar[d]^{h_2} & P \ar[r] \ar[d]^{\varphi } & 0 \\ 0 \ar[r] & M_1 \ar[r]^{f_1} & M_2 \ar[r]^{f_2} & M_3 \ar[r] & 0 . }$

By (4) there is a map $k_1: R^ m \to M_1$ such that $k_1 \circ g_1 = h_1$ . Now define $h'_2: R^ m \to M_2$ by $h_2' = h_2 - f_1 \circ k_1$ . Then

$h'_2 \circ g_1 = h_2 \circ g_1 - f_1 \circ k_1 \circ g_1 = h_2 \circ g_1 - f_1 \circ h_1 = 0 .$

Hence by passing to the quotient $h'_2$ defines a map $\varphi ': P \to M_2$ such that $\varphi ' \circ g_2 = h_2'$ . In a diagram, we have

$\xymatrix{ R^ m \ar[r]^{g_2} \ar[d]_{h'_2} & P \ar[d]^{\varphi } \ar[dl]_{\varphi '} \\ M_2 \ar[r]^{f_2} & M_3. }$

where the top triangle commutes. We claim that $\varphi '$ is the desired lift, i.e. that $f_2 \circ \varphi ' = \varphi$ . From the definitions we have

$f_2 \circ \varphi ' \circ g_2 = f_2 \circ h'_2 = f_2 \circ h_2 - f_2 \circ f_1 \circ k_1 = f_2 \circ h_2 = \varphi \circ g_2.$

Since $g_2$ is surjective, this finishes the proof.

Now we show (5) implies (6). Write $M_{3}$ as the colimit of a directed system of finitely presented modules $M_{3, i}$ , see Lemma 10.11.3. Let $M_{2, i}$ be the fiber product of $M_{3, i}$ and $M_{2}$ over $M_{3}$ —by definition this is the submodule of $M_2 \times M_{3, i}$ consisting of elements whose two projections onto $M_3$ are equal. Let $M_{1, i}$ be the kernel of the projection $M_{2, i} \to M_{3, i}$ . Then we have a directed system of exact sequences

$0 \to M_{1, i} \to M_{2, i} \to M_{3, i} \to 0,$

and for each $i$ a map of exact sequences

$\xymatrix{ 0 \ar[r] & M_{1, i} \ar[d] \ar[r] & M_{2, i} \ar[r] \ar[d] & M_{3, i} \ar[d] \ar[r] & 0 \\ 0 \ar[r] & M_{1} \ar[r] & M_{2} \ar[r] & M_{3} \ar[r] & 0 }$

compatible with the directed system. From the definition of the fiber product $M_{2, i}$ , it follows that the map $M_{1, i} \to M_1$ is an isomorphism. By (5) there is a map $M_{3, i} \to M_{2}$ lifting $M_{3, i} \to M_3$ , and by the universal property of the fiber product this gives rise to a section of $M_{2, i} \to M_{3, i}$ . Hence the sequences

$0 \to M_{1, i} \to M_{2, i} \to M_{3, i} \to 0$

split. Passing to the colimit, we have a commutative diagram

$\xymatrix{ 0 \ar[r] & \mathop{\mathrm{colim}}\nolimits M_{1, i} \ar[d]^{\cong } \ar[r] & \mathop{\mathrm{colim}}\nolimits M_{2, i} \ar[r] \ar[d] & \mathop{\mathrm{colim}}\nolimits M_{3, i} \ar[d]^{\cong } \ar[r] & 0 \\ 0 \ar[r] & M_{1} \ar[r] & M_{2} \ar[r] & M_{3} \ar[r] & 0 }$

with exact rows and outer vertical maps isomorphisms. Hence $\mathop{\mathrm{colim}}\nolimits M_{2, i} \to M_2$ is also an isomorphism and (6) holds.

Condition (6) implies (1) by Example 10.82.2 (2). $\square$

Comments (4)

Comment #1395 by Fred Rohrer on April 09, 2015 at 21:16

In "(2) $\Rightarrow$ (3)", the basis of $R^m$ should be renamed, e.g. to $(e_j)$ , since $f_1$ is already used.

Comment #1396 by Fred Rohrer on April 09, 2015 at 21:18

In "(5) $\Rightarrow$ (6)", at the end of the first sentence, insert ", see Lemma 00HA".

Comment #1397 by Fred Rohrer on April 09, 2015 at 21:22

In "(5) $\Rightarrow$ (6)", one basically makes use of (the special case for modules of) 08N3 and 05QB, appearing only later. Maybe insert the module versions somewhere earlier?

Comment #1412 by Johan on April 15, 2015 at 18:37

@Fred: I fixed your first two comments #1395 and #1936.

About #1397: We need somebody to add a bit more to the section on the snake lemma in algebra.tex and in it discuss these types of lemmas and a tiny bit about exact functors between module categories, or more generally some categories whose objects are abelian groups with a bit of extra structure (but not formalize this notion because that is just aweful). And then point out that the notions agree with those in categories.tex via the material in homology.tex. I have added this to the todo list.

See here.

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