# Proof sketch/verification - proving a pullback in a diagram is an equalizer.

We're given the following diagram: We assume the category has products and equalizers. Given $$E, p_1, p_2$$ as a pullback here, we seek to prove that $$E,e$$ form an equalizer of $$f \circ \pi_1$$ and $$g \circ \pi_2$$.

(The assumption that the category has products and the use of $$A \times B$$ as an object comes from this being the converse of a theorem if you're at all curious.)

I'm mostly wanting to check whether the following would prove that.

(Note: I'm not providing the full, formal proof. This is more a sketch proof; in this post, I'm mostly checking whether I understand what I need to prove, rather than outright proving it. I probably can prove all of these myself, I just want to make sure that these lead to the desired conclusion, if that makes any sense. I'm also a little lost at the very end of the proof as I note when appropriate.)

So what I think I need to prove and the order in which I do:

• First, we assume that $$E, p_1, p_2$$ is a pullback in the given diagram.

• Through this assumption, we prove $$(f \circ \pi_1) \circ e$$ = $$(g \circ \pi_2) \circ e$$, one of the properties of an equalizer.

• We introduce a second pullback $$Z, z_1, z_2$$ with arrow $$z : Z \rightarrow A \times B$$ given by the pullback and unique, and establish the same property as before, i.e. $$(f \circ \pi_1) \circ z$$ = $$(g \circ \pi_2) \circ z$$

At this point I'm a bit lost. I'm debating over which I need to show...

• ... there exists a unique arrow $$u : Z \rightarrow E$$ ...
• ... or that the triangle in the diagram below commutes, i.e. $$e \circ u = z$$ I feel like both might be easy enough to prove with the assumption that equalizer arrows, i.e. $$e$$, are monomorphisms, though I also feel that is too much of an assumption given we haven't proven $$E,e$$ are really an equalizer at this point yet.

I guess my sticking point is invoking the universal property. Part of me is also saying I can invoke it right now - since we have $$E, e$$ and $$Z, z$$ equalizing the pair of arrows - and just establish the arrow $$u$$ exists and is unique and makes the triangle commute immediately.

Anyone have any ideas?

• Start by figuring out what you have and where to go from there. You have a pullback, a product, and an arrow $z : Z\to A\times B$ (which, by the product rule, you can decompose uniquely). You want a unique arrow $u$ such that $e \circ u = z$. – Larry B. Oct 18 '18 at 0:45
• If you want to show that $(E,e)$ is an equalizer of $f\pi_1$ and $g\pi_2$, you don't want to introduce a second pullback into the proof. You want $z$ to be an arbitrary arrow with $f\pi_1z=g\pi_2z$. – Malice Vidrine Oct 18 '18 at 1:02
• @LarryB. what exactly do you mean by decomposing $z$ uniquely? I'm not sure how to go about that. – Eevee Trainer Oct 18 '18 at 3:15
• The decomposition in question comes from the universal property of the product $A\times B$; there's a bijective correspondence between pairs $z_1:Z\to A$, $z_2:Z\to B$, and morphisms $z:Z\to A\times B$, with $z_1=\pi_1z$ and $z_2=pi_2z$. The key observation is that if $f\pi_1z=g\pi_2z$, this is exactly a commuting square of the sort with respect to which the pullback $E$ is universal. – Malice Vidrine Oct 18 '18 at 4:07

## 1 Answer

Given your first diagram, if you want to show that $$e:E\to A\times B$$ is an equalizer of $$f\pi_1$$ and $$g\pi_2$$, it suffices to show that for any $$z:Z\to A\times B$$ with $$f\pi_1z=g\pi_2 z$$, there is exactly one morphism $$r:Z\to E$$ with $$er=z$$.

So assume we have such a $$z$$. Note that such a $$z$$ means we have a commutative square of the sort that $$(E,p_1,p_2)$$ is universal for. Therefore there is a unique morphism $$r:Z\to E$$ with $$p_1r=\pi_1z$$ and $$p_2r=\pi_2z$$. Note that this hasn't said the right thing about $$e$$ (or anything about $$e$$) yet. To get that, we need to use our product $$A\times B$$.

In that first diagram, $$e$$ is the unique map such that $$\pi_1e=p_1$$ and $$\pi_2e=p_2$$. Using these identities in the above equations describing $$r$$, the equality $$p_1r=\pi_1 z$$ becomes $$\pi_1er=\pi_1z,$$ and similarly $$p_2r=\pi_2z$$ becomes $$\pi_2er=\pi_2z.$$ By the universal property of products, though, there is exactly one map $$q:Z\to A\times B$$ with $$\pi_1q=\pi_1z$$ and $$\pi_2q=\pi_2z$$ --- that is to say, $$z$$. And since the map $$er$$ also satisfies this, we have $$er=z$$.

And just in case it's not obvious that $$r$$ being the unique arrow in the pullback diagram implies that it's unique in the equalizer diagram, note that if $$ej=z$$, then $$\pi_nej=\pi_nz$$, and so $$p_nj=\pi_nz$$ ($$n=1,2$$). But by the universal property of the pullback, there can only be one morphism that satisfies those last equations, so $$j=r$$.