# Relative cell complexes in the undercategory $B/\mathscr{M}$ are relative cell complexes in $\mathscr{M}$ - why must we also assume $B$ is a complex?

$$\newcommand{\M}{\mathscr{M}}\newcommand{\I}{\mathcal{I}}\newcommand{\J}{\mathcal{J}}$$The context: the following is claimed in J. May and Ponto's "more concise algebraic topology".

We have some model category $$\M$$ which is generated by arrow classes $$\I,\J$$. For any $$B\in\M$$, the undercategory $$B/\M$$ has a model structure generated by the classes $$B/\I,B/\J$$ which are defined to be the classes of all arrows in the image $$((-)\sqcup B)(\I,\J)$$ respectively. Here, $$(-)\sqcup B:\M\to B/\M$$ is the left adjoint to the forgetful $$U:B/\M\to\M$$.

May and Ponto want to demonstrate that when $$\M$$ is compactly or cofibrantly generated, then so is $$B/\M$$. To do so, they claims that, when $$B$$ is an $$\I,\J$$-cell complex, then relative $$B/\I,B/\J$$-cell complexes in $$B/\M$$ descend (via $$U$$) to relative $$\I,\J$$-cell complexes in $$\M$$.

I don't see why the assumption that $$B$$ is a cell complex is needed.

First of all, by some set theoretic shenanigans I can take all my relative cell complexes to be simple. I also only need to show the result for, say, $$\J$$. A $$B/\J$$-relative cell complex will look like some transfinite sequence $$(X_\alpha)_{\alpha<\lambda}$$ where $$X_{\alpha+1}$$ is obtained from $$X_\alpha$$ as a pushout (in $$B/\M$$) of some $$C\sqcup B\overset{j\sqcup1}{\longleftarrow}A\sqcup B\to X_\alpha$$ (again, a diagram in $$B/\M$$) where $$j:A\to C$$ is an arrow in $$\J$$. This pushout may as well be computed in $$\M$$, however. Colimits in an over category are computed in $$\M$$.

But it's easy to check $$A\hookrightarrow A\sqcup B\to C\sqcup B,\,A\to C\hookrightarrow C\sqcup B$$ defines a pushout square. We can compose pushout squares to find a pushout square $$A\to X_{\alpha}\to X_{\alpha+1},A\to C\to X_{\alpha+1}$$. Since the left leg of this square is an arrow in $$\J$$, continuing in this way finds $$X_\ast$$ as a $$\J$$-relative cell complex. The "middle man" $$B$$ can be omitted entirely.

Am I mistaken, or are the authors?

• Actually, when you slice under an object you have an undercategory. Feb 5 at 14:41
• @ZhenLin Whoops, I messed up the terminology. I thought that $B$ being over $\mathscr{M}$ entails an overcategory Feb 5 at 15:17
• Cross-posted Feb 7 at 18:50

I think both you and the book are right: the forgetful functor does preserve cell complexes, as you show, but the book doesn’t claim otherwise. Specifically, I presume you’re referring to Remark 15.3.7 in May & Ponto:

[…] $$\newcommand{\M}{\mathcal{M}}\newcommand{\I}{\mathcal{I}}\newcommand{\J}{\mathcal{J}}$$ Provided that $$B$$ itself is a cell complex, $$U : B/\M\to \M$$ carries relative cell complexes to relative cell complexes. In this case, we see using the adjunction (15.3.3) that if $$\I$$ and $$\J$$ are compact or small, then so are $$\I/B$$ and $$\J/B$$. Without the proviso on B, this seems to be false. It follows that if $$B$$ is both an $$\I$$-cell complex and a $$\J$$-cell complex and $$\M$$ is compactly or cofibrantly generated, then so is $$B/\M$$.

The phrasing is certainly a bit unclear, but as I read it, the sentence “Without the proviso on $$B$$, this seems to be false” refers to the compactness/smallness claim, and (from that) the compact/cofibrant generation. So while they do take the assumption on $$B$$ already when they state that the forgetful functor preserves cell complexes, I don’t think they’re yet claiming the assumption is necessary. And I agree with your argument; I don’t think the assumption is necessary for that claim.

A couple of side notes:

• The book is by May and Ponto, not just May! Please don’t be sloppy about authors like that — it’s bad practice for many reasons, not least that younger, female, etc. authors tend to be the ones people forget most often.

• Regarding colimits in undercategories: the general statement is that connected colimits in an undercategory $$B/\M$$ are computed as in $$\M$$ — the forgetful functor from the undercategory creates connected colimits.

• It would have been helpful to give the precise citation to Remark 15.3.7! Giving exact references is always good practice — it’s very little extra work when you’re writing (since you already have the reference open at the page in question) but it’s a significant time-saver for anyone chasing up the reference.

• $\newcommand{\colim}{\operatorname{colim}}$Right, thanks. We want $B/I$ to be small. For any domain object $A\sqcup B$ I want $\colim(B/M)(A\sqcup B,X_\alpha)\to(B/M)(A\sqcup B,\colim X_\alpha)$ to be an isomorphism. Adjunction means this is so iff. $\colim M(A,X_\alpha)\to M(A,\colim X_\alpha)$ is an isomorphism, which is true because $I$ is small. Where did I need $B$ to be a cell complex? Feb 7 at 22:27
• And yes, your side notes are well received Feb 7 at 22:28
• ^ Using the fact that these particular colimits are computable in $M$. I also think May and Ponto made a typo with $I/B,J/B$ being written rather than $B/I,B/J$ since it was already remarked that $I/B,J/B$ are small (independent of any other assumptions) Feb 7 at 22:41
• (apologies if this feels like "changing the goalposts", my confusion has just now shifted to a different place. I still don't know why we need B to be a cell complex) Feb 9 at 0:12
• I think the book meant what I initially thought they meant. Earlier in that remark they say: “ forgetful functor U : M/B -> M clearly carries relative cell complexes to relative cell complexes (both defined starting from either I or J). It follows that if I and J are compact or small, then so are I/B and J/B. Therefore, if M is compactly or cofibrantly generated, then so is M/B.” They feel as I do that, given that $U$ takes relative cell complexes to relative cell complexes, it’s immediate that compactness is inherited. The phrasing suggests they believe U does this for M/B but not for B/M Apr 3 at 11:43

Note that $$B$$ itself is a relative cell complex (the initial one) so if relative cell complexes are absolute cell complexes too then $$B$$ is an absolute cell complex.

• I am not trying to show $B$ is an absolute cell complex, nor even a relative cell complex Feb 7 at 20:29
• I think I misunderstood the question. You’re right, $U$ preserves relative cell complexes regardless of the nature of $B$ Feb 7 at 21:03