# The Unapologetic Mathematician

## Categories with Zero Morphisms

Todd Trimble cleared up the confusion with pointed sets and zero morphisms. The problem is that my sources were wrong to say that the monoidal structure on $\mathbf{pSet}$ is the categorical product. First, here’s the answer in Todd’s own words:

The zero morphism statement is fine as stated. It just
follows from the fact that the standard monoidal product
on pointed sets is smash product.

FWIW, here are details. Recall that the smash product
of two pointed sets A, B (with basepoints 0_A, 0_B) is

A /\ B = A x B / ({0_A} x B) \/ (A x {0_B})

where the indicated quotient means the denominator is
identified with a point (and we define the basepoint of
A /\ B to be this point).

If C is a category enriched in pointed sets, and a and b
are objects of C, we define the zero morphism 0_ab
from a to be to be to be the basepoint of hom(a, b).
Notice that composition

hom(b, c) /\ hom(a, b) –> hom(a, c)

takes a pair (g, 0_ab) [similarly, a pair (0_bc, f)] to 0_ac,
by definition of smash product. Hence, composition of
a zero morphism with another morphism, on either side,
yields a zero morphism.

Truth be told I’d been thinking about these “smash products”, but they aren’t the regular categorical product I’d been assured worked.

Okay, let’s go over this a bit. Given pointed sets $(X,x_0)$ and $(Y,y_0)$ we define their smash product $(X\wedge Y,(xy)_0)$ as follows. First we take everything in $X$ other than $x_0$ and call it $\tilde{X}$. Similarly, we say $\tilde{Y}$ is everything in $Y$ except the point $y_0$. We take the product $\tilde{X}\times\tilde{Y}$ and then throw in the new point $(xy)_0$, which we use as the new special point.

Another way to look at it is to take the regular product $X\times Y$, but to “smash” it down a bit. We say that every pair of the form $(x_0,y)$ or $(x,y_0)$ is “really the same”, and smash all of them together into one special point.

The identity object for this monoidal structure on $\mathbf{pSet}$ is the set $(\{a,b\},a)$, as you should check. Also verify that the smash product is associative (up to isomorphism, naturally).

Now a $\mathbf{pSet}$-category $\mathcal{C}$ has a special morphism in each hom-set, which we’ll (leadingly) call $0:A\rightarrow B$. If we take any other arrow $f:B\rightarrow C$, together they form the pair $(0,f)\in\hom_\mathcal{C}(A,B)\wedge\hom_\mathcal{C}(B,C)$. But since $0\in\hom_\mathcal{C}(A,B)$ is the special point of that set, this pair (and any other pair of the form $(0,g)$ or $(g,0)$ is the marked point of the smash product. Then the composition function has to send this special point to the special point $0\in\hom_\mathcal{C}(A,C)$. Voila: zero morphisms. Compose one with anything and you get another zero morphism back.

August 21, 2007 - Posted by | Category theory

## 1 Comment »

1. Right. Let me add (perhaps it’s already been said) that some of these apparently exotic monoidal products are easily motivated by paying attention to the ‘hom side’ of things first.

For example, for abelian groups B and C, the set of homomorphisms hom(B, C) carries an obvious abelian group structure. Then, given another abelian group A, the tensor product A \otimes B is precisely what you need in order to have an isomorphism hom(A \otimes B, C) ~ hom(A, hom(B, C)), natural in C.

For pointed sets B and C, the set of basepoint-preserving maps hom(B, C) carries an obvious basepoint: the map B –> C which takes everything to the basepoint of C. Then, given another pointed set A, the smash product A /\ B is precisely what you need to have an isomorphism hom(A /\ B, C) ~ hom(A, hom(B, C)), natural in C. This explains “why” the smash product.

In a similar way, you can use this idea to figure out the “right” tensor product for Banach spaces.

For cartesian product on pointed spaces, there is no associated ‘Hom’ [no matter how exotic] for which there is a natural isomorphism hom(A x B, C) ~ hom(A, Hom(B, C)). In other words, the functor — x B has no right adjoint. (If it did, then it would preserve coproducts, but it doesn’t! Why not?) For that reason, cartesian product is not a very good product to use for pointed sets.

Comment by Todd Trimble | August 22, 2007 | Reply