## February 2, 2018

### A Problem on Pushouts and Pullbacks

#### Posted by John Baez

I have a problem involving pullbacks and pushouts. This problem arose in work with Kenny Courser on an application of category theory. But you don’t need to understand anything about that application to understand — and I hope solve! —our problem.

If you can solve it, we will credit you in our paper.

We’re working in the category of finite sets and functions. As you may know, given composable cospans like this:

we can compose them using a pushout and get a new cospan:

going all the way from $S$ to $U$.

Now suppose we have composable maps between cospans, by which I mean commutative diagrams like this:

In our application the vertical arrows are all surjections, so I’m noting that in case it helps. We can compose these maps between cospans and get a new map between cospans like this:

The newly created vertical arrow will again be a surjection.

So far so good. But here’s my problem. I’m also assuming all the squares we start with are pullbacks:

I want to prove that the squares we get are pullbacks:

That’s the problem!

Here’s my feeble attempt to solve it. By symmetry it suffices to prove that either one of the squares we get are pullbacks, say the left one. This is the pasting of two smaller squares:

The left-hand smaller square is a pullback, so it suffices —- though perhaps it’s not necessary — to prove that the right-hand smaller square is a pullback.

To prove that the right-hand smaller square is a pullback, it may help to contemplate this cube:

We know that the vertical arrows are surjections, the top and bottom faces are pushouts, and the two left-most faces are pullbacks. We’re trying to prove that the two right-most faces are pullbacks. By symmetry, it’s enough to prove this for either one.

Believe it or not, this sort of situation shows up in the theory of adhesive categories, of which the category of finite sets is an example. People think about a cube like this, with all assumptions except that the vertical squares may not be epic. And in an adhesive category, if either $o_1'$ or $i_2'$ is monic, we’re done!

That is: when we have the above cube in the category of finite sets, and either $o_1'$ or $i_2'$ is monic, and the top and bottom faces are pushouts, and the two left-most faces are pullbacks, then the the two right-most faces are pullbacks!

So that’s where I stand, paralyzed with fear. I’m trying to reach the same conclusion, with all the same assumptions except without assuming that $o_1'$ or $i_2'$ are monic… but I do know that all the vertical arrows in the cube are epic.

That is: if we have the above cube in the category of finite sets, and the vertical arrows are epic, and the top and bottom faces are pushouts, and the two left-most faces are pullbacks, can you show the two right-most faces are pullbacks???

Kenny has looked in vain for counterexamples, so this might actually be true. Or maybe a somewhat different strategy is required. Perhaps somehow the composite square here is a pullback even though the right-hand part is not:

Help! This sort of category theory is not my strong suit.

Posted at February 2, 2018 1:15 AM UTC

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### Re: A Problem on Pushouts and Pullbacks

I think you can simplify the problem by assuming that $i_1$, $i'_1$, $o_2$ and $o'_2$ are identities.

If the answer to your question is “yes” in general then of course it’s yes in this special case. Conversely, if it’s yes in this special case then it’s yes in general, because the composite of pullback squares is a pullback square (as you observe).

So, you could remove all mention of $S$, $f$, $S'$, $i_1$, $i'_1$, $U$, $h$, $U'$, $o_2$ and $o'_2$ from the question.

Posted by: Tom Leinster on February 2, 2018 3:28 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Typos: in the three paragraphs starting from the reference to adhesive categories, you mention $i'_1$ several times, but I assume you mean $i'_2$.

Posted by: Tom Leinster on February 2, 2018 3:35 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Yes, thanks! I’ll fix that.

Posted by: John Baez on February 2, 2018 4:10 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Oh… that sounds like Distributivity… “adhesive”? Who came up with that?

(I like distributivity-in-the-homotopy-sense because it works for connected spaces and sort-of makes All The Cohomology of Homotopy-Types work…)

A certain wizard once wrote something to the effect that The Thing to Do (to decide if it’s true) is to compare the actual pullbacks with the spaces trying to occupy their niches… the ones you’ve called $X,Y$ … as you describe, one only need do the argument once. (He was describing a construction of Opetopic Sets, I think, but … similar deal)

Is It True?

A thing in the pullback competing with $X$ is an annotated pair $(x':X', v:X+_T Y ; x' = (p+q) v )$ ( $\mathbf{Set}$ has UIP ) and you … you basically want to decide that $v$ actually comes from an $x : X$ with $x'= p x$ that we can pin-down exactly

A-priori, the options are $v \in X - o T$ , $v \in Y - i T$ and $v \in T / (i \sim o)$; but the second case $v\in Y - i T$ is ruled-out because then you’d be able to reduce $(p+q) v = q v \in Y' - i' T'$. If $v \in X - o T$, then … that $v$ is the $x\in X$ we want. The tricky case is $v$ being in the image of $T$.

So, here’s my candidate bad $X,T,Y$ $\array{ & & x_1 & & & & x_2 \\ & \nearrow & & \nwarrow & & & & \nwarrow \\ t_1 & & & & t_2 & & & & t_3 \\ & \searrow & & & & \searrow & & \swarrow \\ & & y_1 & & & & y_2 }$ As you can see, anything in the pushout $X +_T Y$ does come from an $x\in X$, but we can’t tell which one!

Posted by: Jesse C. McKeown on February 2, 2018 3:42 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Something about this is bothering me, though: the question is then whether bad $X,T,Y$ can arise (… without $X'$ knowing how to fix things) when $T$ is also pullback for two cospans with a common foot.

So we have a span $X' \leftarrow T' \rightarrow Y'$. And for every $x' : X'$ we have a set $P x'$ and for every $y' : Y'$ a set $Q y'$; and the pullback of all-the-$P$s over $T'$ is the set of pairs

(1)$( t : T \& p : P o' t)$

while the pullback of all-the-$Q$s over $T'$ is the set

(2)$( t : T \& q : Q i' t)$

But we want these to be equivalent in a way commuting with the map forgetting $p$ or $q$, which means we want, for every $t$, An Isomorphism $\varphi_t : P o' t \sim Q i' t$.

The pushout $X +_T Y$ then looks like this mess:

(3)$\frac{(y : Y' \& Q y) \sqcup (x : X' \& P x)}{(o't,p) \sim (i't,\varphi_t p)}$

which maps to $X' +_{T'} Y'$ by $(x,p) \mapsto x$ and $(y,q) \mapsto y$

I’m still rather expecting that the answer should be “no”, in general, but now it’s because of $X',T',Y'$ that look like this: $\array{ t_1 & \to & x \\ \downarrow & & \uparrow \\ y & \leftarrow & t_2 }$ and the fact that the $\varphi_{t_j}$ don’t have to be related in any way.

Posted by: Jesse C. McKeown on February 2, 2018 4:25 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Can you explain what those emoticons mean, for those of us who learned to speak English in school instead of l33t?

Posted by: Mike Shulman on February 2, 2018 3:49 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Presumably, it’s something like:

You answer this problem for me, and I’ll bow down before you, expressing my deferential respect.

Posted by: David Corfield on February 2, 2018 9:34 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Well, both Mike and Tom already have that on my part, me being the novice in this story. But nevertheless, I appreciate their helpful responses :) (no emoticon pun intended, Mike)

Posted by: Kenny Courser on February 2, 2018 3:17 PM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

To me the first one looks like someone holding up their hands to say “stop”. Or maybe pretending to be moose antlers. And there’s another one at the end that I can’t interpret at all. Partly because they’re both too small for me to see clearly.

I’m sorry if this comes across as whiny or Luddite, and I don’t mean to pick on John particularly here. But I really preferred the Internet back before large sections of it regressed to logographic writing. Can I enter a plea that people go back to using words to say exactly what they mean?

Posted by: Mike Shulman on February 2, 2018 10:47 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

The second one is rolling eyes with a smirk. I’m honestly not quite sure what John meant by that one either. But perhaps it was something like “Oh silly me, I’m really not good at this sort of category theory.”? Something like that.

I’d probably have gone with something like “😅” but that’s kind of an even more modern take. It roughly means “laughs nervously” or something to that effect.

At least if I’m reading John’s intentions right. Could be totally off-base! 😅

Posted by: kram1032 on February 2, 2018 12:39 PM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

It sounds like maybe Mike wasn’t seeing the animations, but only still images? This would certainly make them less clear.

Posted by: Dan Christensen on February 4, 2018 12:20 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

That would explain it. He also mentioned they look small; they look perfectly big enough to me.

Posted by: John Baez on February 4, 2018 3:48 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Here’s the size that I see:

At first I didn’t see any animations, but now I realize that it animates once if I shift-reload the page. This is probably because I’ve set image.animation_mode = once in my Firefox about:config, which I did because I generally find web pages that have constant animations going on them to be quite annoying. I think the same would be true of a constantly animated emoticon in a blog comment: constant motion attracts the eye and would be distracting when I’m trying to read text.

Posted by: Mike Shulman on February 4, 2018 8:58 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

You can right-click on the image and view image properties. This shows the image file name, which in this case indicates that the emoticons are “bowing down” and “roll eyes”.

But I know that misses your point. Personally, emoticons and emojis don’t bother me, though I don’t always understand what they mean and sometimes have to resort to manoeuvres like the one just described. I don’t use them myself beyond the primitive :-) and variants, which I think are very useful in internet communications where tone can be hard to convey.

Posted by: Tom Leinster on February 2, 2018 2:19 PM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Yes, I use the basic (-: and sometimes (-:O too, for that reason.

Posted by: Mike Shulman on February 2, 2018 3:53 PM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Here’s what they mean:

Bowing down to thank anyone who helps me solve this problem, in humorously exaggerated praise:

Nervously contemplating my weakness as a category theorist when it comes to reasoning with mixtures of limits and colimits:

Their meaning, like facial expressions, is context-dependent. If you don’t get them, you’re only missing a decorative ‘frosting’ on the cake of my post. I don’t use emojis to convey crucial steps in a mathematical argument.

Communication has always changed with the technology that enables it, and that’s not going to stop. Communication has always been ambiguous, too, especially to those not “in the know” (whatever that happens to mean). The subtle academic in-jokes of category theorists are just as hard for nonexperts to interpret as emoticons.

Everyone gets grumpy when they feel they’re not “in the know”, and most people, past a certain age, feel that the communication habits of young people are sloppy and standards are declining.

We all know category theorists who consider the practice of discussing math on blogs and forums to be dangerously decadent. Going back much further, in the Phaedrus Socrates warned of the dangers of writing as opposed to dialogue: the reader can’t ask questions of the text, and the text is addressed to everyone, not crafted to the individual. The interesting thing about blogs and smart phones is that they turn writing back into a form of dialogue. Emoticons add a primitive version of facial expressions… which will probably become more sophisticated as time goes by.

Posted by: John Baez on February 2, 2018 5:27 PM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Thanks for the explanations, John.

For what it’s worth, it seems to me that “Communication has always changed”, “Communication has always been ambiguous”, and “most people, past a certain age, feel that the communication habits of young people are sloppy and standards are declining” are fallacious arguments. The fact that change always happens has no bearing on whether a particular change is good or not; the fact that ambiguity is always present has no bearing on whether a particular ambiguity is good or not; and the fact that older people always complain about things has no bearing on whether a particular complaint is justified. (The third one could be described as an ad populum reversal or as a Bulverism. The first is related to an appeal to novelty though not I think quite the same. I’m not sure what to call the second.)

Whenever I make a category theorist’s in-joke I generally do try to explain it, being aware that there are plenty of novice category theorists (and even non-category-theorists) who read this blog, and in my experience you are generally quite good at doing the same. And that’s regarding something that is explicitly the subject matter of this blog.

As for facial expressions, I would say rather that emoticons add another channel of communication that is “intended to carry some of the same meaning as facial expressions”. Note that reading actual facial expressions correctly is already hard; indeed even adults not infrequently do it wrong. And that’s something we’ve practiced nearly every waking moment of our lives, and are probably even biologically hard-wired for. If each emoticon had a standard meaning that I could look up in a dictionary, that would be one thing (although I don’t quite know how I would even in principle go about “looking up an animated gif in a dictionary” at a technical level), but that doesn’t seem to be the case. That’s why I think it is better to stick to a very small number of such symbols which have well-established meanings that are broadly applicable, like :) for “joke”. I already try to avoid :P for “sticking tongue out” because it’s not clear at who one is sticking the tongue out; apparently the meaning of :P has been the subject of a court case.

Posted by: Mike Shulman on February 4, 2018 9:51 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Mike wrote:

For what it’s worth, it seems to me that “Communication has always changed”, “Communication has always been ambiguous”, and “most people, past a certain age, feel that the communication habits of young people are sloppy and standards are declining” are fallacious arguments.

They weren’t arguments; they were just observations. I didn’t feel any need to justify my communication methods. I was just ruminating about how people have had discussions of this general sort for millennia.

Posted by: John Baez on February 5, 2018 4:03 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Here, I believe, is a counterexample.

$X' = Y' = 1$

$T' = \{0,1\}\quad X = \{x,y\}\quad Y = \{u,v\} \quad T = \{a,b,c,d\}$

$g(a)=g(b) = 0 \quad g(c)=g(d)=1$

$i_2(a) = i_2(c) = x \quad i_2(b)=i_2(d)=y$

$o_1(a)=o_1(d) = u \quad o_1(b)=o_1(c)=v$

The vertical arrows $g,p,q$ (the latter two defined uniquely) are surjections. The two pullback squares $T = X\times_{X'} T'$ and $T = Y \times_{Y'} T'$ exhibit $4 = 2\times 2$ in two different ways. But both pushouts $X' +_{T'} Y' = X +_T Y$ are a 1-element set, so that $X$ and $Y$ are not the pullbacks $1\times_1 1 = 1$.

Posted by: Mike Shulman on February 2, 2018 5:19 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Nice, good catch! Thanks Mike, Tom, and Jesse, and anyone else who put at least an ounce of thought into this.

Posted by: Kenny Courser on February 2, 2018 3:04 PM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Here’s how I came up with this, by the way. This statement is true without any assumptions of monicity or epicity for homotopy pushouts and pullbacks of homotopy spaces (or more generally in any $\infty$-topos). So, I thought, let’s find an example of such a cube where $X,Y,T,X',Y',T'$ are all sets but the (homotopy) pushouts are not; then probably it will no longer work if we replace the homotopy pushouts by their set-truncations (the pushouts in $Set$).

One of the simplest (perhaps the simplest) pushout of sets that is not a set is

$\array { 2 & \to & 1 \\ \downarrow && \downarrow \\ 1 & \to & S^1 }$

exhibiting $S^1$ as the suspension of $2 = S^0$. Let’s put that on the bottom as $X',Y',T'$. Now we need a map $P\to S^1$ whose pullback over $1\to S^1$ is a set. The obvious choice is $P=1$, in which case this pullback computes the loop space $\Omega S^1 = \mathbb{Z}$. This leads to a counterexample with $X=Y=T=\mathbb{Z}$, $g(t) = t \mod 2$, $i_2(t) = \lfloor \frac{t}{2}\rfloor$, and $o_1(t) = \lceil \frac{t}{2}\rceil$.

But this doesn’t live in finite sets, so instead we can let $P=S^1$ and $P\to S^1$ be the double covering map (the first example being equivalently the universal cover $R\to S^1$). This leads to the example I described above.

Posted by: Mike Shulman on February 2, 2018 3:54 PM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Wow — thanks, Mike! It’s especially cool how you got the counterexample by thinking homotopically.

This makes me much more comfortable about demanding that the legs of our cospans

be monic. Kenny and I are working on a theory where these cospans, equipped with some extra structure, are ‘open Markov processes’: sets of states equipped with time-dependent probability distributions where probability can flow in or out of the cospans’ feet. We’re building a symmetric monoidal double category where the horizontal 1-cells are open Markov processes and the 2-cells are ‘coarse-grainings’, diagrams like this:

again equipped with extra structure. The idea of a coarse-graining is that it maps an open Markov process to a simpler one. Coarse-grainings are already widely studied for ordinary Markov processes, but we need to define them for open Markov processes.

We don’t need the cospans’ legs to be monic when working solely with the horizontal 1-cells. But we need the squares in our 2-cells to be pullbacks to get certain things to work… and thanks to you, we now know that for this property to be preserved by horizontal composition, we need the legs of our cospans to be monic.

This is a mild imposition, but I’m happier knowing that we’re making it because we need it.

In your honor, I will not express my gratitude with an emoticon.

Posted by: John Baez on February 2, 2018 6:30 PM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

When you see something like this, you feel pity for the category of sets.

Posted by: Joachim Kock on February 11, 2018 1:40 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Did you feel a similar pity back here?

Posted by: David Corfield on February 12, 2018 5:07 PM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

(My remark was to the counter example.)

Posted by: Joachim Kock on February 11, 2018 1:45 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

The pushouts in Set that are Van Kampen have been characterized by Michael Löwe in his technical report with the title “Van-Kampen Pushouts for Sets and Graphs”, including many pictures and examples.

Posted by: Tobias Heindel on February 2, 2018 10:55 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

Tobias’ comment points to the original result, which exactly characterises Van Kampen Pushouts in the cat. of Sets.

A more general result for Van Kampen Colimits in more general categories has been proved by Uwe Wolter and me: https://arxiv.org/abs/1710.09784

Posted by: Harald König on February 2, 2018 11:25 AM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

I came too late to this to look for my own counterexample, but was figuring that any statement like this (in the category of finite sets) either has an obvious proof, or is false. Is that unfair?

Posted by: Allen Knutson on February 4, 2018 1:43 PM | Permalink | Reply to this

### Re: A Problem on Pushouts and Pullbacks

So the true statement that I wound up needing to use in my paper is this:

Given a cube in the category of finite sets:

where the top and bottom faces are pushouts, the two left vertical faces are pullbacks, and the arrows $o_1'$ and $i_2'$ are monic, then the two right vertical faces are pullbacks.

Actually a stronger statement is known: if either $o_1'$ or $i_2'$ is a monic, the bottom square is a pushout, and the two left vertical faces are pullbacks, then the top face is a pushout iff the two right vertical faces are pullbacks.

I suppose some people think the proof of this is obvious, but it doesn’t feel obvious to me. And it seems to me that the best proofs sneak up on this statement through some abstract ideas, rather than ‘bashing it out’.

Posted by: John Baez on February 5, 2018 1:14 AM | Permalink | Reply to this

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