# The Unapologetic Mathematician

## You cannot run from Carnival!

It’s up and running at MathNotations. And even running is a kind of dance!

## Multiple products, coproducts, equalizers, coequalizers, pullbacks, pushouts…

We’ve got products and coproducts of two objects at a time, equalizers and coequalizers of two morphisms at a time, and pushouts and pullbacks of two objects at a time over a third. We can generalize all of these.

First let’s do equalizers. If we have three arrows $f_1$, $f_2$, and $f_3$ from an object $A$ to an object $B$, the equalizer of all three will be an arrow $e:E\rightarrow A$ so that $f_1\circ e=f_2\circ e=f_3\circ e$. If we have regular equalizers we can build something to satisfy this property.

Let $e_{1,2}:E_{1,2}\rightarrow A$ be the equalizer of $f_1$ and $f_2$. That is, $f_1\circ e_{1,2}=f_2\circ e_{1,2}$. Now we have $f_1\circ e_{1,2}$ and $f_3\circ e_{1,2}$ going from $E_{1,2}$ to $B$, so we can take their equalizer $e_{1,2,3}:E_{1,2,3}\rightarrow E_{1,2}$. Now $f_1\circ e_{1,2}\circ e_{1,2,3}=f_3\circ e_{1,2}\circ e_{1,2,3}$. And clearly $f_1\circ e_{1,2}\circ e_{1,2,3}=f_2\circ e_{1,2}\circ e_{1,2,3}$. So we have $e_{1,2}\circ e_{1,2,3}:E_{1,2,3}\rightarrow A$ as the equalizer of all three morphisms.

Of course we didn’t have to start with $f_1$ and $f_2$. We could have started with $f_1$ and $f_3$, taken their equalizer, and so on to get another equalizer $e_{1,3}\circ e_{1,3,2}:E_{1,3,2}\rightarrow A$. It’s important to note here that these two equalizers are not the same in general. Whatever category we’re working in will have some construction to give binary equalizers, and when we apply it twice in two different ways we’ll usually get two different results. But the two results are isomorphic, since each is a couniversal object in the category of arrows that equalize the three arrows we started with, and couniversal objects are unique up to isomorphism. Speaking roughly it’s not too much of a problem to talk about “the” equalizer, but it’s useful to keep in the back of your mind that we’re really talking about an isomorphism class of such objects.

We can similarly define equalizers of any finite number of parallel arrows, and if we have equalizers of pairs we have all of them. We can even define the equalizer of an infinite family of parallel arrows, though now we can’t show that they exist using only pairwise equalizers. And, of course, all this goes the same for coequalizers.

Now we’ll do multiple products. Pullbacks are a kind of product so they’ll come along for the ride, and pushouts and coproducts are dual.

If we have three objects $A_1$, $A_2$, and $A_3$, their product $P=\prod\limits_{i=1}^3A_i$ will be an object with arrows $\pi_i:P\rightarrow A_i$ that is couniversal among such objects. Again, if we have binary products we can define the product of three objects: $(A_1\times A_2)\times A_3$ has arrows $\pi_{A_1}\circ\pi_{A_1\times A_2}$, $\pi_{A_2}\circ\pi_{A_1\times A_2}$, and $\pi_{A_3}$ that go from the product to the three objects, and it’s straightforward to verify that it satisfies the universal property required. Again, we can also take the product $A_1\times(A_2\times A_3)$ with the obvious arrows, and this also satisfies the universal property. And again, these two may not be the same, but the universal property guarantees that they’re isomorphic.

We can do the same thing to define the product of any finite number of objects, or even an infinite family (though now we can’t build them from finite products). One interesting case is when we take the product of no objects. This is just an object $T$, since we don’t need any projection arrows out of it to the factors. For any other object $A$ (with no particular arrows out) there is a unique arrow from $A$ to $T$. That is, the product of no objects is a terminal object. Similarly, the coproduct of no objects is an initial object.

June 15, 2007 Posted by | Category theory | 1 Comment