Lemma 12.5.11. Let $\mathcal{A}$ be an abelian category. Let

$\xymatrix{ w\ar[r]^ f\ar[d]_ g & y\ar[d]^ h\\ x\ar[r]^ k & z }$

be a commutative diagram.

1. The diagram is cartesian if and only if

$0 \to w \xrightarrow {(g, f)} x \oplus y \xrightarrow {(k, -h)} z$

is exact.

2. The diagram is cocartesian if and only if

$w \xrightarrow {(g, -f)} x \oplus y \xrightarrow {(k, h)} z \to 0$

is exact.

Proof. Let $u = (g, f) : w \to x \oplus y$ and $v = (k, -h) : x \oplus y \to z$. Let $p : x \oplus y \to x$ and $q : x \oplus y \to y$ be the canonical projections. Let $i : \mathop{\mathrm{Ker}}(v) \to x \oplus y$ be the canonical injection. By Example 12.5.6, the diagram is cartesian if and only if there exists an isomorphism $r : \mathop{\mathrm{Ker}}(v) \to w$ with $f \circ r = q \circ i$ and $g \circ r = p \circ i$. The sequence $0 \to w \overset {u} \to x \oplus y \overset {v} \to z$ is exact if and only if there exists an isomorphism $r : \mathop{\mathrm{Ker}}(v) \to w$ with $u \circ r = i$. But given $r : \mathop{\mathrm{Ker}}(v) \to w$, we have $f \circ r = q \circ i$ and $g \circ r = p \circ i$ if and only if $q \circ u \circ r= f \circ r = q \circ i$ and $p \circ u \circ r = g \circ r = p \circ i$, hence if and only if $u \circ r = i$. This proves (1), and then (2) follows by duality. $\square$

Comment #380 by Fan on

Typo: the canonical injection is $i: \ker(v) \to x \oplus y$, not to $x \oplus v$.

I'm afraid that there is a sign error somewhere, as the sequence is not even a chain if I take $w = x = y = z$ and $f = g = h = k = 1$.

Comment #386 by on

@#380: Fixed this and most of your other comments. See here. Thanks very much!

Comment #3843 by Junho Won on

I think the second exact sequence is missing a 0 on the left.

Comment #3844 by Laurent Moret-Bailly on

@Junho Won: no. For instance, if $x=y=z=0$, the diagram is cocartesian for any $w$.

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