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?

Question

$\newcommand{\M}{\mathscr{M}}\newcommand{\I}{\mathcal{I}}\newcommand{\J}{\mathcal{J}}$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?

Answer 1

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.

Answer 2

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.