The equalizer of the projections is the diagonal map

Question

I want to prove that the equalizer of $\pi_0,\pi_1: B^2 \rightarrow B$ is the diagonal $\delta_B:B \rightarrow B^2$.

That $\delta_B$ equalizes is easy enough but the universal part is a little bit more tricky for me. Let $x:T \rightarrow B^2$ such that $\pi_0x =\pi_1 x$, then there should exist a $\bar{x}:T \rightarrow B$ such that $x=\delta_B \bar{x}$.

My first thought was to try $\bar{x}=\pi_0 x$, but that doesn't help since $\delta_B \bar{x}= \delta_B (\pi_0 x) = (\delta_B \pi_0)x$ doesn't equal $x$.

Any help is much appreciated!

Answer 1

By definition $\delta_B$ is the unique arrow $B\to B\times B$ with $\pi_0\delta_B=1_B=\pi_1\delta_B$.

If $x:T\to B\times B$ such that $\pi_0x=\pi_1x$ then for $\bar x:=\pi_0x=\pi_1x$ we find the equalities:

• $\pi_0(\delta_B\bar x)=(\pi_0\delta_B)\bar x=1_B\bar x=\bar x=\pi_0x$
• $\pi_1(\delta_B\bar x)=(\pi_1\delta_B)\bar x=1_B\bar x=\bar x=\pi_1x$

This allows the conclusion that: $$\delta_B\bar x=x$$

Answer 2

$\delta_B \pi_0 x$ does equal $x$.

Consider that it has the properties $$ \pi_0 (\delta_B \pi_0 x) = \pi_0 x \qquad \pi_1 (\delta_B \pi_0 x) = \pi_0 x$$ because $\pi_0\delta_B = \pi_1\delta_B = \mathit{id}_B$.

This means that $\delta_B \pi_0 x$ satisfies the universal property required of the map $\pi_0 x \times \pi_0 x$. But so does $x$ itself -- and by definition of $B\times B$ there is a unique map that does this.

Answer 3

I want to prove this in a different style via the notion of a universal element. My particular proof has the benefit of not needing to know the answer beforehand, but the other approaches could easily be adapted to calculate the resulting arrow as well. I'm not proposing that this is a better way of solving this particular problem, I just think it's worth seeing what it would look like. I do think a universal element/representability approach is generally a good way to deal with universal properties though.

Finding the equalizer of $\pi_0,\pi_1$ means finding an object that represents the functor $\{h\mid h\in\text{Hom}(-,B\times B)\land\pi_0 \circ h = \pi_1 \circ h\}$, that is we want an object $E$ and a natural isomorphism $$\text{Hom}(-,E) \cong \{h\mid h\in\text{Hom}(-,B\times B)\land \pi_0 \circ h = \pi_1 \circ h\}$$ The universal property for $B\times B$ looks like the following in this form. $$\text{Hom}(-,B\times B)\cong\text{Hom}(-,B)\times\text{Hom}(-,B)$$ The universal element, in general, is the image of $id$ via the natural transformation. So for products, the pair of arrows $(\pi_0,\pi_1)$ is the universal element. For pairs, I'll write the inverse natural transformation as $(p,q)\mapsto \langle p,q \rangle$. That it is the inverse means $\langle \pi_0,\pi_1 \rangle = id$ which is one of the universal properties for pairs. $\delta = \langle id,id \rangle$.

Okay. The above was all context. Here's the actual calculation. $$\begin{align} \text{Hom}(-,E) & \cong \{h\mid h\in\text{Hom}(-,B\times B)\land\pi_0 \circ h = \pi_1 \circ h\} \\ & \cong \{(p,q)\mid (p,q)\in\text{Hom}(-,B)\times\text{Hom}(-,B)\land p = q\} \\ & = \{ (p,p) \mid p\in\text{Hom}(-,B)\} \\ & \cong \text{Hom}(-,B) \end{align}$$

By Yoneda, $E \cong B$, but we'll just choose $E = B$. To calculate the universal element, we just push $id$ through the natural isomorphisms backwards. $id \mapsto (id, id) \mapsto \langle id,id \rangle = \delta$. $\square$

The first isomorphism is by definition of $E$ being an equalizer. The second comes from representability of pairs, but it may not be completely obvious, so you should work out why it holds. The remainder is set-theoretic reasoning. In general, representability reduces categorical reasoning to set-theoretic reasoning allowing you to leverage the mathematical experience you already have. Slightly less generally, via representability limits and colimits are (both) calculated from limits in $\mathbf{Set}$. More precisely, they are limits in a category of $\mathbf{Set}$-valued functors, but those are calculated point-wise.