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Stream: theory: category theory

Topic: pushout gap maps in a commutative cube

Chaitanya Leena Subramaniam (Jan 19 2026 at 13:59):

I realised something interesting over the weekend.

Given a commutative cube in a category with pushouts
image.png

let $f$ and $g$ be the pushout gap maps of the rear face (ABXY) and front face (A'B'X'Y') respectively, and let $h$ and $k$ be the pushout gap maps of the left face (ABA'B') and right face (XYX'Y') respectively. Then a neat pasting argument (which works equally well in an $\infty$-category) shows that the pushout gap map of the square $f\to g$ (i.e. the square between the front and back pushout gap maps) is isomorphic to the pushout gap map of the square $h\to k$ (the square between the left and right pushout gap maps).

Is this already known to anyone here? And if so, do you have a nice conceptual explanation?

(N.B. The pushout gap map of the face ABXY is the map $f$ below)
image.png

Morgan Rogers (he/him) (Jan 19 2026 at 17:53):

That statement doesn't immediately type check to me. $f,g$ have codomains $Y,Y'$ respectively, whereas $h,k$ have codomains $B',Y'$. It doesn't seem likely that $B' \cong Y$ in general, so which things are being connected by isomorphisms here?

Chaitanya Leena Subramaniam (Jan 19 2026 at 18:11):

The statement is that the (domains of the) pushout gap maps of the following squares are isomorphic
image.png
image.png

I agree that it's not obvious, but a clever choice of pushout pasting shows that they are.

David Wärn (Jan 19 2026 at 21:55):

I think of this as saying that the colimit of "the cube minus the terminal vertex (Y')" can be computed by taking iterated pushouts in the manner you describe. So the sources of these pushout products are isomorphic, because they're all the colimit of cube minus terminal vertex

Chaitanya Leena Subramaniam (Jan 20 2026 at 06:23):

I agree. Moreover, the fact that the order doesn't matter makes it seem like there is some module/algebra structure at play.

Chaitanya Leena Subramaniam (Jan 20 2026 at 06:36):

I think I’ve got it.

The external pushout-product is a bi-cocontinuous functor $\mathcal{S}^\to\times\mathcal{C}^\to \to \mathcal{C}^\to$, and gives $\mathcal{C}^\to$ an $\mathcal{S}^\to$-module structure (where the internal pushout-product gives $\mathcal{S}^\to$ a commutative monoid structure in the symmetric monoidal category of cocomplete categories and cocontinuous functors --- here $\mathcal{S}$ is the category of spaces, or sets if we're talking about 1-categories).

The associated cocontinous functor $\mathcal{S}^\to\otimes\mathcal{C}^\to = (\mathcal{C}^\to)^\to \to \mathcal{C}^\to$ is the functor that computes the pushout gap map of a square. Then the module structure on $\mathcal{C}^\to$ is what gives us that the gap map of a cube can be computed in two equivalent ways.

David Wärn (Jan 20 2026 at 10:13):

You can also think of the pushout gap map as left Kan extension along $\mathrm{max} : [1]^2 \to [1]$. Since $\mathrm{max}$ is associative, ternary maximum $[1]^3 \to [1]$ factors as $[1]^2 \times [1] \xrightarrow{\mathrm{max} \times [1]} [1] \times [1] \xrightarrow{\mathrm{max}} [1]$, and so does LKE along ternary maximum. This should give the desired result.

Morgan Rogers (he/him) (Jan 20 2026 at 11:56):

Aha, I had missed that there was one more pushout operation involved. So it's an isomorphism between domains of two maps into $Y'$, and I follow @David Wärn 's argument for why those should be the same.
I don't know why you specify "two equivalent ways", though @Chaitanya Leena Subramaniam . There are clearly three!

Chaitanya Leena Subramaniam (Jan 20 2026 at 12:20):

@Morgan Rogers (he/him) : You're right, I framed my sentence wrong -- I wanted to say "these two ways of computing the gap map are equivalent" :upside_down:

@David Wärn : I think the two arguments are the same -- $\mathrm{max}$ is the coproduct on $[1]$, and its Day convolution gives the pushout-product on $\mathcal{S}$. Then, unwinding the passage between bi-cocontinuous functors and cocontinuous functors out of the tensor product in cocomplete categories gives your argument. Interestingly, this should also work for Day convolutions associated to other symmetric monoidal diagram categories.

Morgan Rogers (he/him) (Jan 21 2026 at 08:55):

I wasn't criticising the phrasing, I just thought it was worth pointing out the further symmetry of the situation :sweat_smile: