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I am reading May's 'More Concise Algebraic Topology' and I ran into something I couldn't figure out.

Let $(X,*)$ be a based CW-complex. The category of $X$ is said to be less than $k$ if the diagonal map $\Delta: X \to X^{k+1} = X \times X \times ... \times X$ is homotopic, in the based sense, to a map $X \to F_kX:=(* \times X \times ... \times X) \cup (X \times * \times ... \times X) \cup ... \cup (X \times X \times ... \times *)$, i.e. $F_k X$ consists of those points $(x_1,...,x_{k+1})$ such that one of the $x_i$ is the based point $*$.

Meanwhile, a stratification of $X$ is a filtration by subcomplexes $*=Y_0 \subset Y_1 \subset ... \subset Y_k=X$ such that all the cells of $Y_i$ have boundary in $Y_{i-1}$, e.g. $Y_i=i^{th}$ skeleton of $X$ is a stratification.

May claims in his book that if $X$ has a stratification of length $k$, then the category of $X$ is less than $k$, but I am having trouble providing a proof for his claim.

I can prove the statement when $k=1$: $Y_1$ is just a wedge of spheres and using cellular approximation theorem, we can homotope the diagonal map $\Delta: S^n \to S^n \times S^n$ to a map in $S^n \bigvee S^n$. I am assuming one proceeds upwards the stratification using similar techniques. As a start, by homotopy extention property, we now have a map $Y_2 \to Y_2 \times Y_2$ such that $Y_1$ is mapped to $* \times Y_1 \cup Y_1 \times *$. Then for each cell in $Y_2$, it is attached by pasting along $Y_1$, but trying to invoke cellular approximation theorem here again cannot eliminate the possibility that a part of $Y_2$ will be mapped to $Y_1 \times Y_1$, which is not inside our desired range. Of course I'm not using the extra copy of $X$ here, and it clearly should be the key. But I can't see a way to apply cellular approximation along this extra copy to get what I want.

Any hints or solutions are appreciated!

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  • $\begingroup$ The extra copy is key. Whitehead (Elements of Homotopy Theory) proves that the mapping cone of a map $X\to Y$ has category $\le$ the category of $Y$ $+1$. From there, you get your result. $\endgroup$ – Justin Young Oct 23 '17 at 20:12
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Here is a sketch of the argument. First, if $X\simeq Y$ then $\text{cat} X = \text{cat} Y$. This is relatively easy.

Next result, if $f:X\to Y$ is any map then $\text{cat} Mf \le \text{cat} Y + 1$, where $Mf$ is the mapping cone of $f$.

To prove this, first you replace $X\to Y$ by a cofibration, so WLOG you can assume that $X\to Y$ is a cofibration. In this case, $X\to CX$, $Y\to Mf$ and $CX\to Mf$ are also cofibrations. Now, you can extend the homotopy on $Y$ to get a map $Mf \to Mf^k$ that is homotopic to the diagonal, and so that $Y$ lands in $F_k Y$ (we assume here that $\text{cat} Y \le k$). Next, you can use the fact that $CX$ is contractible to get a homotopy from the identity to a map $Mf \to Mf$ that sends $CX$ to the basepoint. Now, you take the product of these constructed maps, precomposed with the diagonal to get $Mf \to Mf \times Mf \to Mf^k \times Mf = Mf^{k+1}$, a map that lands in $F_{k+1}Mf$ by construction, and is homotopic to the diagonal map $Mf\to Mf^{k+1}$.

To get your result, note that you get from $Y_i$ to $Y_{i+1}$ by taking the mapping cone of $X\to Y_i$ where $X$ is a wedge of spheres (working in basepoint world here).

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  • $\begingroup$ Very clear answer, thank you! Also sorry for the late upvote+accept! $\endgroup$ – Tsang Oct 27 '17 at 22:56

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