Lemma 4.19.4 (09WQ)—The Stacks project

Lemma 4.19.4. Let $\mathcal{I}$ be an index category, i.e., a category. Assume that for every pair of objects $x, y$ of $\mathcal{I}$ there exist an object $z$ and morphisms $x \to z$ and $y \to z$. Then

If $M$ and $N$ are diagrams of sets over $\mathcal{I}$, then $\mathop{\mathrm{colim}}\nolimits (M_ i \times N_ i) \to \mathop{\mathrm{colim}}\nolimits M_ i \times \mathop{\mathrm{colim}}\nolimits N_ i$ is surjective,
in general colimits of diagrams of sets over $\mathcal{I}$ do not commute with finite nonempty products.

Proof. Proof of (1). Let $(\overline{m}, \overline{n})$ be an element of $\mathop{\mathrm{colim}}\nolimits M_ i \times \mathop{\mathrm{colim}}\nolimits N_ i$. Then we can find $m \in M_ x$ and $n \in N_ y$ for some $x, y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ such that $m$ maps to $\overline{m}$ and $n$ maps to $\overline{n}$. See Section 4.15. Choose $a : x \to z$ and $b : y \to z$ in $\mathcal{I}$. Then $(M(a)(m), N(b)(n))$ is an element of $(M \times N)_ z$ whose image in $\mathop{\mathrm{colim}}\nolimits (M_ i \times N_ i)$ maps to $(\overline{m}, \overline{n})$ as desired.

Proof of (2). Let $G$ be a non-trivial group and let $\mathcal{I}$ be the one-object category with endomorphism monoid $G$. Then $\mathcal{I}$ trivially satisfies the condition stated in the lemma. Now let $G$ act on itself by translation and view the $G$-set $G$ as a set-valued $\mathcal{I}$-diagram. Then

\[ \mathop{\mathrm{colim}}\nolimits _\mathcal {I} G \times \mathop{\mathrm{colim}}\nolimits _\mathcal {I} G \cong G/G \times G/G \]

is not isomorphic to

\[ \mathop{\mathrm{colim}}\nolimits _\mathcal {I} (G \times G) \cong (G \times G)/G \]

This example indicates that you cannot just drop the additional condition Lemma 4.19.2 even if you only care about finite products. $\square$

Comments (5)

Comment #1043 by Bas Edixhoven on September 18, 2014 at 16:01

The proof given shows that colim commutes with finite products, but the lemma says nonempty products.

Comment #1047 by Johan on September 21, 2014 at 14:25

OK, thanks! Two remarks to see that the lemma is now sharp:

I think the lemma does not hold for the empty product because if $\mathcal{I}$ is empty, then $\colim_\mathcal{I} \{*\} = \emptyset$ whereas $\prod_{\emptyset} \colim_\mathcal{I} anything = \{*\}$ .

I think the lemma does not hold for infinite products because $\prod_m \bigcup_n [0, n] \not = \bigcup_n \prod_m [0, n]$ for example.

The change is here.

Comment #2640 by Steffen Sagave on July 10, 2017 at 11:39

The statement of this lemma appears to be wrong.

Here is a counterexample: Let $G$ be a non-trivial group and let $\mathcal{I}$ be the one-object category with endomorphism monoid $G$ . Then $\mathcal{I}$ trivially satisfies the condition stated in the lemma. Now let $G$ act on itself by translation and view the $G$ -set $G$ as a set-valued $\mathcal{I}$ -diagram. Then $\mathrm{colim}_{\mathcal{I}} (G) \times \mathrm{colim}_{\mathcal{I}} (G) \cong G/G \times G/G$ is not isomorphic to $\mathrm{colim}_{\mathcal{I}}(G \times G) \cong (G \times G)/G\ .$

This example indicates that you cannot just drop the additional condition of tag 002W even if you only care about finite products.

Comment #2641 by Johan on July 10, 2017 at 18:49

OK, yes, this is just terrible! I will say more: it is a disgrace! Thanks very much for pointing this out. Will fix this right away.

Comment #2642 by Johan on July 10, 2017 at 20:45

Thanks again for pointing this out. Luckily we didn't use this lemma anywhere but in this section and in one totally unimportant spot. Actually, the lemmas 4.19.4, 4.19.5, 4.19.6, and 4.19.7 should probably all be moved to the obsolete chapter as they now appear to me to be totally useless (and indeed never get used). I'll leave them here for a while and perhaps move them later.

The changes are here. In about 30 minutes the changes will be online.

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5 comment(s) on Section 4.19: Filtered colimits

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