# Doubt about proof of factorization $f=pi$, where $i$ is acyclic cofibration and $p$ is fibration

I try to understand a proof in More Concise Algebraic Topology: Localization, completions and model categories by May & Ponto (pdf). The proof is on page 262, and it is for the statement

Any map $f:X\to Y$ factors as the composite of an acyclic cofibration and a fibration

The authors proceed as follows. Let $Z_0=X$ and $\rho_0=f$. We assume inductively that we have constructed a map $\rho_n:Z_n\to Y$. We define $N\rho_n$ as the pullback of $Y^I\to Y\leftarrow Z_n$. Where the first map is $\varepsilon_0$ (evaluation at $0$). The composite $N\rho_n\to Y^I\to Y$ can also be written as $N\rho_n\to N\rho_n\times I\to Y$ where the first map is inclusion at $0$ and the second map is the adjunct of the map $N\rho_n\to Y^I$. If we now define $Z_{n+1}$ as the pushout of $N\rho_n\times I\leftarrow N\rho_n\to Z_n$, this induces a unique map $\rho_{n+1}:Z_{n+1}\to Y$ making certain triangles commute. For the first $n$, I have displayed the various arising maps in the following diagram. I hope it's not too confusing, the arrows must be thought of as evolving towards the front for increasing $n$.

Let $Z$ be the colimit of the sequence $Z_0\hookrightarrow Z_1\hookrightarrow \dots$, with $\nu:Z_0\hookrightarrow Z$ the initial inclusion into this colimit. Since all the maps $\nu_n$ are acyclic cofibrations, so is $\nu$. If $\rho:Z\to Y$ denotes the map induced by the $\rho_n$, then $\rho\nu=f$, so it suffices to show that $\rho$ is a fibration. This is the case if in the following square there is a lift, where $N\rho$ is the pullback of the diagram $Y^I\to Y\leftarrow Z$.

One can show that, since we are working in the category of weak Hausdorff $k$-spaces, the canonical map $\text{colim}N\rho_n\to N\rho$ is a homeomorphism. Also, $\text{colim}(N\rho_n\times I)\cong N\rho\times I$. This way, the map $N\rho\to Z$ is the colimit of the maps $N\rho_n\to Z_n$. So practically all the objects in the square are colimits. Now according to the authors, the diagonal is induced by the maps $\lambda_n:N\rho_n\to Z_{n+1}$. Here comes the point I am doubtful about: I don't see why the square with $\lambda_0$ and $\lambda_1$ commutes.

In fact, I'm pretty sure it does not commute: A point in $N\rho_0\times I$ is of the form $(q,z, t)$, which by the map $\lambda_0$ is sent to its equivalence class and then via $\nu_1$ to the class represented by $\left(\overline{q(t)},(q,z,t),0\right)$ (where the bar denotes constant path) while $(q,z,t)$ is sent via the other map to $\left(q,\left(\overline{\rho_0(z)},z,0\right),t\right)$ and then via $\lambda_1$ to that element's equivalence class in $Z_2$.

Is there any way how one could use the constructions of pushouts and pullbacks to acquire the desired lift?

• You might want to try linking this in the homotopy theory chat for math overflow. – PVAL-inactive Mar 7 '15 at 2:36

The issue you noticed is an artifact of the fact that May and Ponto seem to have gotten their proof from Cole's 2006 paper "Many homotopy categories are homotopy categories." In Barthel and Riehl's 2013 paper "On the construction of functorial factorizations for model categories," it is noted that Williamson noticed this error, and in (5.5) and §6.1 Barthel and Riehl point out the same issue you do: Cole's construction does not even work for two steps!

If I am correct in thinking that you want a proof for the statement on p. 343 of More concise algebraic topology, where the statement is for the the category of compactly generated (weak Hausdorff) spaces, with the model structure consisting of Hurewicz fibrations and cofibrations with weak equivalences being homotopy equivalences, then I think Strøm's original proof from his 1972 paper "The homotopy category is a homotopy category" goes through. The construction is follows (this is summarized in §2 of Barthel and Riehl).

First recall that the mapping path space $Nf$ of $f\colon X \to Y$ is defined to be the pullback: $$\require{AMScd} \begin{CD} Nf @>>> Y^I\\ @VVV @VVV\\ X @>{f}>> Y \end{CD}$$ and that as stated in May's A concise course in algebraic topology, p. 48, we have a factorization $$X \overset{j}{\longrightarrow} Nf \overset{\pi}{\longrightarrow} Y$$ where $j$ is a homotopy equivalence and $\pi$ is a fibration. The issue is that $j$ is not necessarily a cofibration, so Strøm factors $j$ through a space $E$ which is the pushout $$\begin{CD} X \times (0,1] @>{j \times \mathrm{id}}>> Nf \times (0,1]\\ @VVV @VVV\\ X \times I @>>> E \end{CD}$$ From this pushout you obtain a map $E \to Nf \times I$ using the maps $Nf \times (0,1] \to Nf \times I$ and $j \times \mathrm{id}\colon X \times I \to Nf \times I$. We can then factor $j$ as $$X \overset{i_0}{\longrightarrow} X \times I \longrightarrow E \longrightarrow Nf \times I \longrightarrow Nf$$ One can then show the composition $i \colon X \to E$ of the first two maps is an acyclic cofibration and that the composition $\pi'\colon E \to Nf$ is a fibration, using Theorems 8 and 9 from Strøm, "Note on cofibrations, II." Then, $p = \pi \circ \pi'$ is a fibration, and so $X \overset{i}{\to} E \overset{p}{\to} Y$ is the factorization desired.

To fix Cole's construction that you originally asked about in your question, refer to the rest of Barthel and Riehl's paper; there the statement is Thm. 5.22.

EDIT 1: May is aware of the issue as he noted in a MathOverflow answer. In particular, an outline version of the proof of the factorization is given on p. 1121 of the 2014 paper "Six model structures for DG-modules over DGAs" by Barthel, May, and Riehl.

EDIT 2: Here is a proof that $\pi'\colon E \to Nf$ is a fibration. We show this by verifying the RLP directly: consider the diagram $$\begin{CD} Z @>g>> E\\ @V{i_0}VV @VV{\pi'}V\\ Z \times I @>G>> Nf \end{CD}$$ We want to find a diagonal $\overline{G}\colon Z \times I \to E$ making the diagram commute. Defining $$\overline{G}(z,t) = (G(z,t),t + (1-t)\operatorname{pr}_Ig(z))$$ where $\operatorname{pr}_I\colon E \to I$ is that induced by the pushout diagram for $E$ using the second projection maps $Nf \times (0,1] \to I$ and $X \times I \to I$, we have $$(\overline{G} \circ i_0)(z) = \overline{G}(z,0) = (G(z,0),\operatorname{pr}_I g(z)) = g(z)$$ and $$(\pi' \circ \overline{G})(z,t) = \pi'(\overline{G}(z,t)) = \pi'(G(z,t),t + (1-t)\operatorname{pr}_Ig(z)) = G(z,t)$$ where I think it's important to remember that $E$ is just $X \times I$ glued to $Nf \times (0,1]$ along $j \times \operatorname{id}$.

Note this is in the proof of Prop. 1 in Strøm. I'd also like to mention that May and Sigurdsson in Parametrized homotopy theory reproduce Strøm's result as Thm. 4.4.4, and it looks to me that it doesn't rely on the (same) incorrect proof due to Cole which is Thm. 4.4.2.

• Thanks, this is really helpful. It seems like Strøm proved that $i:X\to E$ is an acyclic cofibration by showing that $E$ deformation retracts to $X\times I$ and then to $X$ (which holds because $N_f$ deformation retracts to $j(X)$), and then using the fact that $X$ is also a zero set under the obvious map $E\to I$. – Stefan Hamcke Mar 10 '15 at 4:20
• I have trouble proving that $E\to N_f$ is a fibration. Strøm only says that it is one without giving further insight. Do you know how to prove it or do you have a reference for that? – Stefan Hamcke Mar 13 '15 at 3:54
• @StefanHamcke I've provided the proof that $E \to N_f$ is a fibration (I think you don't get notified so I thought I'd comment). – Takumi Murayama Mar 15 '15 at 6:21
• Thanks. But it seems that $E$ must be given the topology as a subspace of $N_f\times I$, which is what both Strøm and May/Sigurdsson do. Otherwise, the map from $Z\times I$ to $E$ need not be continuous. I'm going to check whether everything works with the coarser topology. – Stefan Hamcke Mar 17 '15 at 3:50
• Yes, $E$ with the subspace topology still deformation retracts to $j(X)\times I$ and then to $X$, and the maps $E\to I$ and $E\to N_f$ are still continuous (Well, they are just projections). Actually, I had thought of such a lift $Z\times I\to E$ myself, but couldn't show its continuity. But now I see that the coarser topology does the job. – Stefan Hamcke Mar 17 '15 at 4:06