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I'm trying to complete the following pattern

product  : monoidal product : coproduct
pullback :         ?        : pushout

That is, if the monoidal product is a common generalization of a product and a coproduct, what is the common generalization of a pullback and pushout? The question I'm asking can naturally be extended to other sorts of categorical (co)limits as well, such as (co)equalizers.

We can also look at this from another angle. A product in the slice category $\mathcal C / X$ is a pullback in the underlying category $\mathcal C$. A coproduct in the coslice category $X / \mathcal C$ is a pushout in the underlying category $\mathcal C$. It feels like it should be possible to fill in the pattern for the monoidal case, but I don't know how.

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    $\begingroup$ A monoidal category such that all its slices/coslices inherit a monoidal structure? Hmm.. $\endgroup$
    – David Roberts ♦
    Commented Dec 21, 2021 at 0:04
  • $\begingroup$ Maybe it is a 2-category (or a variation on such concept, such as a bicategory): products are pullbacks over a prescribed object, namely the terminal one, whereas a monoidal category is bicategory with a single object. If we want products over a variety of objects (=pullbacks), the "monoidal counterpart" would be a bicategory with many objects. In short, pullbacks define the bicategory of spans. $\endgroup$
    – D.-C. Cisinski
    Commented Dec 21, 2021 at 12:58
  • $\begingroup$ Conclusion: the notion of pullback is an instance of a composition law for $1$-cells in a bicategory. This specializes to cartesian products as a monoidal product when we restrict to the point. $\endgroup$
    – D.-C. Cisinski
    Commented Dec 21, 2021 at 17:11
  • $\begingroup$ What archetypal examples do you have in mind? If you're willing to consider symmetric monoidal instead of monoidal structure, I can point you to local independent (co) products, which seem to be a useful axiomatisation of fibred symmetric monoidal structure that I learned about recently: coalg.org/mfps-calco2017/mfps-papers/6-simpson.pdf $\endgroup$
    – vikraman
    Commented Dec 21, 2021 at 23:05
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    $\begingroup$ My answer over here is in principle applicable here, too, but I'm not sure what exactly the conclusion ends up being. $\endgroup$
    – Tim Campion
    Commented Dec 27, 2021 at 17:05

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