# Proving the pullback of monics is monic.

I mostly just want to double-check my reasoning in my proof. For clarity's sake, the diagram we are working with is

where $$m$$ is monic, the top-left corner is a pullback, and we wish to show that $$m'$$ is monic.

Here's a sketch of my proof. I can provide more thorough reasoning if wanted, I'm just providing the rough outline of my reasoning for brevity's sake. (I was able to prove each individual thing well enough to satisfy myself, I just want to be sure the overall reasoning itself is all right.)

1. First, we begin by considering another object $$X$$ with parallel arrows $$a,b : X \rightarrow M'$$ such that $$m' \circ a = m' \circ b$$

2. Before even beginning the proof proper, we note the end goal: to prove $$m'$$ is monic. Our construction helps with that, as will be seen. The goal is to show $$a=b$$ when $$m' \circ a = m' \circ b$$. The goal, then, is to invoke the universal property of the pullback: in doing so, we assure there exists a unique arrow $$u : X \rightarrow M'$$. Because of this, it follows $$a = b = u$$. And thus $$m'$$ is monic, because our assumption $$m' \circ a = m' \circ b$$ led to $$a = b$$.

3. Through our assumptions from the construction of $$X, a, b$$ and from the facts that $$m$$ is monic and the original diagram is a pullback, we next need to prove the commuting of the diagram when X is in play. A rough look at it:

(Slight digression)

It's not pretty to look at - I should really learn how to do arrows in LaTeX - so for clarification's sake, the essential equalities we need to verify are:

$$g\circ a=g\circ b\\ m'\circ a=m'\circ b\\ m\circ (g\circ a)=f\circ (m'\circ a)\\ m\circ (g\circ b)=f\circ (m'\circ b)$$

The first is the commuting of the top trapezoid, the second of the left one, and the third and fourth assure the commuting of the outer square.

1. With the necessary commutings assured, we can invoke the universal property of the pullback. Thus, $$\exists ! u : X \rightarrow M'$$, etc. etc.

2. Since $$u$$ is unique, $$a = b = u$$. Since our assumption $$m' \circ a = m' \circ b$$ implied $$a = b$$, alongside the other assumptions (pullback, $$m$$ monic) then $$m'$$ is monic.

For what it's worth, I believe this is essentially the same line of reasoning communicated in this question. I'm mostly just elaborating further on it, sort of explaining it to myself if you will, since it didn't 100% make sense to me when I saw it there.

• Possible duplicate of Pullbacks of monic morphisms. – Shaun Oct 15 '18 at 23:26
• No, wait; this seems to be a proof-verification. – Shaun Oct 15 '18 at 23:26
• I'm not sure about your 3rd paragraph. You only write four formulas and explain their meaning, you don't prove them. So it's actually not a proof. But do you suggest it as a proof? I don't quite understand what do you mean by "reasoning". – Oskar Oct 15 '18 at 23:29
• This is all correct. I might emphasize that the equation $g\circ a=g\circ b$ is the one that follows from $m$ being monic. – Berci Oct 15 '18 at 23:29
• @Oskar I'm not actually proving them in this question. I'm mostly saying "does this line of reasoning make sense," i.e. "if I prove all of these things, does it properly follow $m'$ is monic?" I can prove all of these things - I just want to make sure that the things I'm proving actually lead to the conclusion I want, if that makes any sense. – Eevee Trainer Oct 15 '18 at 23:30

The bones of your proof are good, and you know where you want to go with it. You also know how to prove this, and you are familiar with the universal properties of pullbacks and monics. Your technical knowledge is spot-on. All that your proof needs is a more clear exposition.

In particular, the four equalities you gave are the crux of the proof. If you expand that part, it makes the proof more clear, particularly how you are using the properties of monic $$m$$ to prove $$m^\prime$$ is monic (which you allude to in your step 3).

The order of equalities is this:

By assumption: $$m^\prime \circ a = m^\prime \circ b$$

Function application: $$f\circ (m^\prime \circ a) = f \circ (m^\prime \circ b)$$

Pullback square commutes: $$(g\circ a) \circ m = f\circ (m^\prime \circ a) = f \circ (m^\prime \circ b) = (g\circ b) \circ m$$

Monic $$m$$: $$g\circ a = g \circ b$$

Pullback property: Since $$g\circ a = g \circ b$$ and $$m^\prime \circ a = m^\prime \circ b$$, there is a unique arrow $$u$$ such that $$m^\prime \circ u = m^\prime \circ a = m^\prime \circ b$$ (similar with $$g$$ but we don't care so much about that).

Then, exactly as you reasoned in steps 4 and 5, $$a = b = u$$, and $$m^\prime$$ is monic.

I see two areas of improvement. One is knowing where to expound your details and where to elide. This is a difficult thing to do, and it only comes with practice. I hope that Math StackExchange can be a supportive community that helps you as you learn this balance.

The second, which might help with the first, is pointing out commutativity. I have found Category Theory diagrams to be super tricky, since not everything necessarily commutes with everything else. Some arrows that could be the composition of two others may not be. Everything commutes in your diagram, and parallel arrows end up being equal, but that is not necessarily so.

Another great way to improve your writing skills is by reading. You can read a proof to understand what is being proved, and your skill in doing this is clear. Additionally, you can read to understand how the author establishes definitions, invokes assumptions in the hypothesis, and connects these facts with logic and the machinery of the mathematics under study to craft a proof.

• I just want to reiterate that this wasn't going to be the proof I'd submit: like you noted, this is the bones of the proof, and in this post I was focusing moreso on the overarching idea of "what I need to prove," as opposed to formality. Thus why I skipped over proving the commuting, for example; sure, I can prove that and expound upon that, I was just moreso concerned in the post with whether that was how I was supposed to do it or not. Though I do agree that I need to get better with properly wording things since I tend to be verbose, far more than necessary I feel. Still, thanks! ^_^ – Eevee Trainer Oct 16 '18 at 0:17
• Excellent proof, and keep up the good work. – Larry B. Oct 16 '18 at 0:21