Contents

category theory

# Contents

## Idea

The concept of pushout product is a natural kind of pairing operation on morphisms in categories equipped with a pairing operation on objects (e.g. a tensor product) and having pushouts. It sends two morphisms to the universal morphism out of the pushout of the span-diagram they form by pairing their domain objects. Regarding arrows in a category as diagrams with domain the interval category, and giving the interval category the natural monoidal product, this is a kind of Day convolution tensor product with the natural copowering over Set replacing a tensor in the coend.

## Definition

Consider

Assume that $\mathcal{E}_3$ has all pushouts.

###### Definition

Given a pair of morphisms

• $f \colon A \to B$ in $\mathcal{E}_1$

• $g \colon X \to Y$ in $\mathcal{E}_2$,

their pushout product is the morphism in $\mathcal{E}_3$

$f \Box g \;\;\colon\;\; A \otimes Y \overset {A \otimes X} {\amalg} B \otimes X \longrightarrow B \otimes Y$

out of the pushout of the two morphisms, which is induced by the universal property of the pushout by the following commuting diagram (induced, in turn, from the functoriality of the tensor product as a functor on a product category):

$\array{ A \otimes X &\overset{f \otimes id_X}{\longrightarrow}& B \otimes X \\ \mathllap{{}^{id_A \otimes g}}\Big\downarrow && \Big\downarrow\mathrlap{{}^{id_B \otimes g}} \\ A \otimes Y &\underset{f \otimes id_Y}{\longrightarrow}& B \otimes Y \mathrlap{\,.} }$

## Properties

For $\mathcal{C}$ any category and $K\subset Mor(\mathcal{C})$ any class of its morphisms, write $K Inj$ for the $K$-injective morphisms and $K Cof \coloneqq (K Inj)Proj$ for the $K Inj$-projective morphisms.

###### Proposition

Let $\mathcal{C}$ be a symmetric closed monoidal category with finite limits and finite colimits, and let $I_1, I_2\subset Mor(\mathcal{C})$ be two classes of its morphisms.

Then under pushout product $\Box$:

$(I_1 Cof) \Box (I_2 Cof) \subset (I_1 \Box I_2) Cof \,.$
###### Proof

By a little Joyal-Tierney calculus.

###### Remark

In the context of monoidal model category theory, prop. implies that for checking the pushout-product axiom in the case of cofibrantly generated model categories it is sufficient to check it on generating cofibrations.

## Examples

### In Sets

It is instructive to consider the elementary case of pushout-products in Sets with respect to the Cartesian product

$\times \;\colon\; Set \times Set \longrightarrow Set \,.$

For a pair of maps/functions of sets

$\array{ f \; \colon & X &\longrightarrow& X' \\ g \; \colon & Y &\longrightarrow& Y' }$

the relevant pushout diagram is

where the dashed arrow denotes the pushout-product map.

###### Example

The pushout-product of sets may be described as the quotient set

$f \,\widehat{\times}\, g \;\; \simeq \;\; \big\{ (x,y'), \, (x',y) \big\}_{ \bigg/ \Big( \big(x,\,g(y)\big) \,\sim\, \big(f(x),\,y\big) \Big) }$

(where each variable ranges over the set denoted by the corresponding capital symbol).

Moreover, if we denote the equivalence class of $(x,y')$ by $[x,y']$, etc. then the coprojections into the pushout product are given by

$\array{ X' \times Y &\xrightarrow{\;\; q_l\;\;}& f \,\widehat{\times}\, g \\ (x',\,y) &\mapsto& [x',\, y] } \;\;\;\;\;\;\;\;\; \text{and} \;\;\;\;\;\;\;\;\; \array{ X \times Y' &\xrightarrow{\;\; q_r\;\;}& f \,\widehat{\times}\, g \\ (x,\,y') &\mapsto& [x,\, y'] }$

and the pushout-product map itself (the dashed arrow) is given as follows:

(1)$\array{ f \widehat{\times} g &\xrightarrow{\phantom{---}}& X' \times Y' \\ [x',\, y] &\mapsto& \big(x',\, g(y)\big) \\ [x,\, y'] &\mapsto& \big(f(x),\, y' \big) \mathrlap{\,.} }$

More informatively, the fibers of the pushout-product map over any $(x',y') \,\in\, X' \times Y'$ look as follows:

(2)$\big( f \widehat{\times} g \big)_{(x',y')} \;\; = \;\; \left\{ \array{ \ast &\vert& (x',y') \;\in\; im(f) \times im(g) \\ f^{-1}\big(\{x'\}\big) &\vert& y' \;\in\; Y' \setminus im(g) \\ g^{-1}\big(\{y'\}\big) &\vert& x' \;\in\; X' \setminus im(f) } \right.$

where

• $\ast$” denotes the singleton set

• $(-)^{-1}$” denotes forming preimages

• $\setminus$” denotes forming complements.

###### Proof

We discuss in more detail how to obtain the fibers (2).

To see the first case: By the assumption that $x' \,\in\, im(f)$ we find $x_1 \in f^{-1}\big(\{x'\}\big)$ and hence by (1) we find $[x_1,y'] \,\in\, \big(f \widehat{\times} g\big)_{(x',y')}$. But since also $y' \,\in\, im(g)$ there is also $y \in g^{-1}\big(\{y'\}\big)$ and hence

$\big[x_1,\, y'\big] \,=\, \big[x_1,\, g(y)\big] \,=\, \big[f(x_1),\, y\big] \,=\, \big[x',\, y\big] \,.$

This shows that all elements of the form $[x,y']$ in the fiber are in fact equal, and also equal to $[x', y]$, which, by the symmetric argument, are in turn equal for all choices of $y$. Therefore there is one single element in the fiber of the pushout-product map, in this case.

To see the second case: Since $y'$ is not in the image of $g$, by (1) the elements in the fiber can only be of the form $[x,y']$ with $x \in f^{-1}\big(\{x'\}\big)$. None of these is contained in the relation – again by the assumption that $y'$ is not in the image of $g$ – and hence they are all distinct.

The third case works the same way, under swapping $X \leftrightarrow Y$.

###### Example

(Pushout-product of injections of sets)
In the special case that both maps are injections, the fibers appearing in (2) are all (-1)-truncated (either empty sets or singleton sets).

This shows that the pushout-product map of two injections of Sets is itself an injection. The following graphics illustrates this for interval-subsets of the plane: ### In Topological Spaces

###### Example

For $n \in \mathbb{N}$, let

$i_n \;\colon\; S^{n-1}\hookrightarrow D^n$

be the canonical sphere inclusions in Top (the generating cofibrations of the classical model structure on topological spaces). Their pushout product (with respect to cartesian product of topological spaces) is given by addition of indices:

$i_{n_1} \Box i_{n_2} \simeq i_{n_1 + n_2} \,.$

Let moreover

$j_n \coloneqq (id,\delta_0) \;\colon\; D^n \hookrightarrow D^n \times I \,.$

Then

$i_{n_1}\Box j_{n_2} \simeq j_{n_1 + n_2} \,.$
###### Proof

To see this, it is profitable to model n-disks and n-spheres, up to homeomorphism, as n-cubes and their boundaries.

To see the idea of the proof, consider the situation in low dimensions, where one readily sees that

$i_1 \Box i_1 \;\colon\; (\; = \;\cup\; \vert\vert\;) \hookrightarrow \Box$

and

$i_1 \Box j_0 \;\colon\; (\; = \;\cup\; \vert \; ) \hookrightarrow \Box \,.$

Generally, $D^n$ may be represented as the space of $n$-tuples of elements in $[0,1]$, and $S^n$ as the suspace of tuples for which at least one of the coordinates is equal to 0 or to 1.

Accordingly $S^{n_1} \times D^{n_2}$ is the spaces of $(n_1+n_2)$-tuples, such that one of the first $n_1$ coordinates is equal to 0 or 1, and hence

$S^{n_1} \times D^{n_2} \;\cup\; D^{n_1} \times S^{n_2} \;\;\simeq\;\; S^{n_1 + n_2} \,.$

And of course it is clear that $D^{n_1} \times D^{n_2} \simeq D^{n_1 + n_2}$. This shows the first case.

For the second, use that $S^{n_1} \times D^{n_2} \times I$ is contractible to $S^{n_1} \times D^{n_2}$ in $D^{n_1} \times D^{n_2} \times I$, and that $S^{n_1} \times D^{n_2}$ is a subspace of $D^{n_1} \times D^{n_2}$.

###### Remark

The relations in example are the key in proving that the classical model structure on topological spaces (on compactly generated topological spaces) is an enriched model category over itself. See there at topological enrichment for more.

### General

###### Example

(pushout-product with initial morphism is tensor product)
With respect to a bifunctor $\otimes \;\colon\; \mathcal{C} \times \mathcal{C} \longrightarrow \mathcal{C}$ which preserves initial objects $\varnothing$ in, say, the first variable, the pushout-product with an initial morphisms $\varnothing \to X$ is given by $X \otimes (-)$:

$\big( \varnothing \to X \big) \widehat{\otimes} g \;\simeq\; id_X \otimes g \,.$

(Of course, in applications $(-)\otimes(-)$ is typically a closed tensor product which hence even preserves all colimits in each variable separately.)

This is because the defining pushout diagram now looks like this:

where the top row is the unique map on the initial object by assumption whence its pushout to the bottom horizontal morphism is also an identity, as shown.

###### Example

(pushout product with identity is identity morphism)
With respect to any bifunctor $\otimes \,\colon\, \mathcal{C} \times \mathcal{C} \to \mathcal{C}$, the pushout-product with with an identity morphism is an identity morphism:

$f \widehat{\otimes} id \;\simeq\; id \;\;\;\;\;\;\;\;\; \text{and} \;\;\;\;\;\;\;\;\; id \,\widehat{\otimes}\, g \;\simeq\; id \,.$

This is because the defining pushout-diagram, say in the first case, now looks like this: because the top morphism is an identity morphism by assumption, so that also its pushout is given by an identity, as shown.

The analogous argument applies in the other variable.

###### Example

With respect to any bifunctor $\otimes \;\colon\; \mathcal{C} \times \mathcal{C} \longrightarrow \mathcal{C}$ the pushout-product of two split epimorphisms

$\array{ f \,\colon & X &\overset{split}{\twoheadrightarrow}& X' \\ g \,\colon & Y &\overset{split}{\twoheadrightarrow}& Y' }$

is an isomorphism, in that the following commuting square is already a pushout

$\array{ X \otimes Y &\overset{id \times g}{\longrightarrow}& X \otimes Y' \\ \mathllap{{}^{f \otimes id}} \Big\downarrow && \Big\downarrow \mathrlap{{}^{ f \otimes id}} \\ X' \otimes Y &\underset{id \otimes g}{\longrightarrow}& X' \otimes Y' \mathrlap{\,.} }$

###### Proof

Given any cocone under the given span, such as shown in the outer part of the following diagram we need to see that there is a unique dashed morphism $\phi$ making the diagram commute, as shown.

Now, by assumption, we have sections

$\array{ \overline{f} \,\colon & X' &\overset{\phantom{--}}{\hookrightarrow}& X & \text{with} & f \circ \overline{f} \,=\, id \\ \overline{g} \,\colon & Y' &\overset{\phantom{--}}{\hookrightarrow}& Y & \text{with} & g \circ \overline{g} \,=\, id }$

from which we obtain a candidate dashed morphism by setting:

$\begin{array}{l} \mathllap{ \phi \; \coloneqq \; } r \circ (\overline{f} \otimes id) \\ \;=\; r \circ (\overline{f} \otimes id) \circ (id \otimes g) \circ (id \otimes \overline{g}) \\ \;=\; r \circ (id \otimes g) \circ (\overline{f} \otimes id) \circ (id \otimes \overline{g}) \\ \;=\; s \circ (f \otimes id) \circ (\overline{f} \otimes id) \circ (id \otimes \overline{g}) \\ \;=\; s \circ (id \otimes \overline{g}) \,. \end{array}$

This does make the bottom triangle commute

$\begin{array}{l} \phi \circ (id \otimes g) \\ \;=\; r \circ (\overline{f} \otimes id) \circ (id \otimes g) \\ \;=\; r \circ (id \otimes g) \circ (\overline{f} \otimes id) \\ \;=\; s \circ (f \otimes id) \circ (\overline{f} \otimes id) \\ \;=\; s \end{array}$

and analogously for the right triangle

$\begin{array}{l} \phi \circ (f \otimes id) \\ \;=\; s \circ (id \otimes \overline{g}) \circ (f \otimes id) \\ \;=\; s \circ (f \otimes id) \circ (id \otimes \overline{g}) \\ \;=\; r \circ (id \otimes g) \circ (id \otimes \overline{g}) \\ \;=\; r \,; \end{array}$

and given any $\phi$ making the diagram commute, we find that it equals the previous one:

$\begin{array}{l} \phi \\ \;=\; \phi \circ (f \otimes id) \circ (\overline{f} \otimes id) \\ \;=\; r \circ (\overline{f} \otimes id) \,. \end{array}$

Pushout-products are prominently discussed in the context of monoidal model category-theory (where they appear in a pushout-product axiom), and here a key motivation are constructions of symmetric monoidal smash products of spectra. See for instance: