We're given the following diagram:

enter image description here

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. $e \circ u = z$

enter image description here

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?

  • 1
    $\begingroup$ 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 $e \circ u = z$. $\endgroup$ – Larry B. Oct 18 '18 at 0:45
  • 1
    $\begingroup$ 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 $f\pi_1z=g\pi_2z$. $\endgroup$ – Malice Vidrine Oct 18 '18 at 1:02
  • $\begingroup$ @LarryB. what exactly do you mean by decomposing $z$ uniquely? I'm not sure how to go about that. $\endgroup$ – Eevee Trainer Oct 18 '18 at 3:15
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    $\begingroup$ 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. $\endgroup$ – Malice Vidrine Oct 18 '18 at 4:07

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 $j=r$.


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