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A square is cartesian exactly when a short sequence is exact

Statement

Consider a commutative square in an abelian category

Then the square is cartesian if and only if the sequence $0 \to P \overset{(\binom{α}{β})}{\to} X \oplus Y \overset{(f, - g)}{\to} Z$ is exact.

Facts & Assumptions

Given: The displayed commutative square.

[L1]

Pullbacks are defined by the usual universal property (Pullbacks and pushouts as limits and colimits of cospans and spans).

[L2]

In a biproduct, a morphism into $X \oplus Y$ is determined by its two projections, and $i_{X} p_{X} + i_{Y} p_{Y} = 1_{X \oplus Y}$ (Biproduct, On a biproduct, the injections and projections satisfy the identity-sum relation).

[L3]

Exactness of $0 \to P \to X \oplus Y \to Z$ is equivalent to the first map being a kernel of the second (Degenerate exactness criteria, Exact sequence and short exact sequence in an abelian category).

Proof

technique · direct

1.1

Assume the square is cartesian. Then $(f, - g) (\binom{α}{β}) = f α - g β = 0$ . If $u : U \to X \oplus Y$ satisfies $(f, - g) u = 0$ , write $x : = p_{X} u$ and $y : = p_{Y} u$ . Then $f x = g y$ , so the pullback property [L1] gives a unique $t : U \to P$ with $α t = x$ and $β t = y$ . By [L2], this implies $(\binom{α}{β}) t = u$ , so $(\binom{α}{β})$ is a kernel of $(f, - g)$ and the sequence is exact by [L3].

L1L2L3assume-hypalgebra

1.2

Assume the sequence is exact. Then [L3] says $(\binom{α}{β})$ is a kernel of $(f, - g)$ , so the square commutes. Given $x : U \to X$ and $y : U \to Y$ with $f x = g y$ , define $u : = i_{X} x + i_{Y} y$ . By [L2], $(f, - g) u = f x - g y = 0$ , so the kernel property gives a unique $t : U \to P$ with $(\binom{α}{β}) t = u$ . Applying $p_{X}$ and $p_{Y}$ yields $α t = x$ and $β t = y$ , proving the pullback property.

L1L2L3assume-hypconstructalgebra

2.1

Therefore the square is cartesian exactly when the displayed sequence is exact.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, Section 12.5, Lemma 12.5.11(1) (standard reference, not scraped)
David Mehrle, Category Theory, Part III, Lemma 7.17 (standard reference, not scraped)