# The Stacks Project

## Tag 09WQ

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 exists an object $z$ and morphisms $x \to z$ and $y \to z$. Then

1. If $M$ and $N$ are diagrams of sets over $\mathcal{I}$, then $\mathop{\rm colim}\nolimits (M_i \times N_i) \to \mathop{\rm colim}\nolimits M_i \times \mathop{\rm colim}\nolimits N_i$ is surjective,
2. 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 $\mathop{\rm colim}\nolimits M_i \times \mathop{\rm colim}\nolimits N_i$. Then we can find $m \in M_x$ and $n \in N_y$ for some $x, y \in \mathop{\rm Ob}\nolimits(\mathcal{I})$ such that $m$ mapsto $\overline{m}$ and $n$ mapsto $\overline{n}$. 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 $\mathop{\rm colim}\nolimits (M_i \times N_i)$ 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 $$\mathop{\rm colim}\nolimits_\mathcal{I} G \times \mathop{\rm colim}\nolimits_\mathcal{I} G \cong G/G \times G/G$$ is not isomorphic to $$\mathop{\rm colim}\nolimits_\mathcal{I} (G \times G) \cong (G \times G)/G$$ This example indicates that you cannot just drop the additional condition Lemma 4.19.2 even if you only care about finite products. $\square$

The code snippet corresponding to this tag is a part of the file categories.tex and is located in lines 2088–2101 (see updates for more information).

\begin{lemma}
\label{lemma-preserve-products}
Let $\mathcal{I}$ be an index category, i.e., a category. Assume
that for every pair of objects $x, y$ of $\mathcal{I}$
there exists an object $z$ and morphisms $x \to z$ and $y \to z$.
Then
\begin{enumerate}
\item If $M$ and $N$ are diagrams of sets over $\mathcal{I}$,
then $\colim (M_i \times N_i) \to \colim M_i \times \colim N_i$
is surjective,
\item in general colimits of diagrams of sets over $\mathcal{I}$
do not commute with finite nonempty products.
\end{enumerate}
\end{lemma}

\begin{proof}
Proof of (1). Let $(\overline{m}, \overline{n})$
be an element of $\colim M_i \times \colim N_i$.
Then we can find $m \in M_x$ and $n \in N_y$ for some
$x, y \in \Ob(\mathcal{I})$ such that $m$ mapsto
$\overline{m}$ and $n$ mapsto $\overline{n}$. See
Section \ref{section-limit-sets}.
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 $\colim (M_i \times N_i)$
maps to $(\overline{m}, \overline{n})$ as desired.

\medskip\noindent
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
$$\colim_\mathcal{I} G \times \colim_\mathcal{I} G \cong G/G \times G/G$$
is not isomorphic to
$$\colim_\mathcal{I} (G \times G) \cong (G \times G)/G$$
This example indicates that you cannot just drop the additional
condition Lemma \ref{lemma-directed-commutes}
even if you only care about finite products.
\end{proof}

Comment #1043 by Bas Edixhoven on September 18, 2014 a 4:01 pm UTC

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

Comment #1047 by Johan (site) on September 21, 2014 a 2:25 pm UTC

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

1. I think the lemma does not hold for the empty product because if $\mathcal{I}$ is empty, then $\mathop{\rm colim}\nolimits_\mathcal{I} \{*\} = \emptyset$ whereas $\prod_{\emptyset} \mathop{\rm colim}\nolimits_\mathcal{I} anything = \{*\}$.
2. 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 (site) on July 10, 2017 a 11:39 am UTC

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 (site) on July 10, 2017 a 6:49 pm UTC

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 (site) on July 10, 2017 a 8:45 pm UTC

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 09WQ, 09WR, 09WS, and 09WT 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.

There are also 3 comments on Section 4.19: Categories.

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