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nLab uses the following definition of van Kampen colimits --- a colimit in a category $\mathbb{C}$ is called van Kampen iff it is preserved by the internal indexing functor $\mathbb{C}/(-) \colon \mathbb{C}^{op} \rightarrow \mathbf{Cat}$ defined as: $$X \mapsto \mathbb{C}/X$$ $$X \overset{f}\rightarrow Y \mapsto \mathbb{C}/Y \overset{f^*}\rightarrow \mathbb{C}/X$$ where $f^*$ is the pullback-along-$f$ functor.

My question is --- where does this definition come from and why are such colimits called "van Kampen"?

In case of coproducts one may notice that the property of being van Kampen in the above sense is equivalent to the usual property of being extensive.

On the other hand, van Kampen pushouts in the above sense do not match the usual definition of van Kampen pushouts from the definition of an adhesive category. For example, if $\mathbb{C} = \mathbf{Set}$ then the internal indexing functor $\mathbf{Set}/(-)$ is equivalent to the usual exponential functor $\mathbf{Set}^{(-)}$, which (like every exponential functor) preserves all colimits; but not every pushout in $\mathbf{Set}$ is van Kampen (pushouts along monomorphisms are).

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    $\begingroup$ arxiv.org/abs/1101.4594 ? $\endgroup$ Jun 22, 2015 at 19:37
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    $\begingroup$ The exponential functor $\mathbf{Set}^{(-)}$ does not send colimits of sets to 2-limits of categories. For example, consider the coequalizer of the diagram $*\rightrightarrows *$. $\endgroup$ Jun 22, 2015 at 21:06
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    $\begingroup$ In fact, the only locally presentable category in which all colimits are van Kampen is the terminal category, since such a category must be an $\infty$-topos. $\endgroup$ Jun 22, 2015 at 21:11
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    $\begingroup$ Maybe your notion of 2-limit is different. According to the nLab definition ncatlab.org/nlab/show/2-limit, the inclusion Set → Cat does not preserve 2-colimits: the 2-colimit of $*\rightrightarrows *$ in Cat is the groupoid $B\mathbb{Z}$. Hence, the 2-limit of $Set\rightrightarrows Set$, where both arrows are the identity, is the category of $\mathbb{Z}$-sets. $\endgroup$ Jun 22, 2015 at 23:30
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    $\begingroup$ @MarcHoyois It appears to me that Michal R. Przybylek is talking about 2-(co)limits in the classical sense of ($\mathbf{Cat}$-)enriched category theory, whereas you are talking about bi(co)limits. $\endgroup$
    – Zhen Lin
    Jun 23, 2015 at 14:55

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Here's a summary of the comments above (which does not answer the question on the origin of the term, that I have no idea).

A colimit in a category $C$ with pullbacks is van Kampen if the indexing functor $C/(-): C^{op} \to Cat$ transforms it into a weak 2-limit, or bilimit, or homotopy limit. These are sometimes called 2-limit or simply limit, but the latter also have stricter meanings which do not give the correct definition.

This is a very strong property: a locally presentable category (or even a locally presentable $(n,1)$-category for any finite $n$) in which all small colimits are universal and all pushouts are van Kampen is necessarily the terminal category, because it must be an $\infty$-topos.

For $C=Set$, for example, the coequalizer of $*\rightrightarrows *$ is not van Kampen, because the weak 2-limit of $Set\rightrightarrows Set$ is the category of $\mathbb{Z}$-sets rather than the category of sets. A colimit in $Set$ is van Kampen iff its category of elements is simply connected. Pushouts have this property if one of the legs is injective, although that's not necessary.

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  • $\begingroup$ Can you elaborate on that last paragraph? I don't see how the category of elements for a coproduct is simply connected. What functor are we even talking about? $\endgroup$ Feb 2, 2023 at 17:04
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    $\begingroup$ I misspoke, the condition is that every connected component of the category of elements should be simply connected. The category of elements of a functor $F: C\to Set$ is the category of pairs $(c,x)$ with $x\in F(c)$. The homotopy type $X$ of this category is the colimit of $F$ in spaces. So the colimit of $F$ in $Set$ is van Kampen iff $Fun(X,Set)=Fun(\pi_0(X),Set)$, which is true iff every component of $X$ is simply connected. $\endgroup$ Feb 2, 2023 at 18:50

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