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The homotopy dimension $h \dim X$ of a space $X$ is the minimal dimension of a CW-complex $Z$ homotopy equivalent to $X$.

I am interested in the generalisation to maps $f\colon X\to Y$. Here's what I think it should be:

The homotopy dimension of $f\colon X\to Y$ is the smallest $k$ such that $f$ factorises through a $k$-dimensional CW-complex up to homotopy (meaning there is a $Z$ of dimension $k$ and maps $g\colon X\to Z$ and $h\colon Z\to Y$ with $f\simeq h\circ g$).

Question 1: Is this the "correct" generalisation? My hesitance stems from the fact that, with this definition, $h\dim 1_X$ (the homotopy dimension of the identity map $1_X\colon X\to X$) does not necessarily equal $h\dim X$. Indeed, the former is the smallest dimension of a CW-complex which dominates $X$, and (I believe) Wall has shown that there are spaces for which $h\dim 1_X$ $<$$\infty$ while $h\dim X=\infty$.

Question 2: I am sure that this is a well-known and well-studied notion, and that I am merely using the wrong search terms. Where should I look in the literature to learn more about this concept?

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up vote 8 down vote accepted

Regarding Question 1: No, I do not think that's correct. In my opinion, the definition should be one of the following:

The relative homotopy dimension of $f: X \to Y$ is $\le k$ if and only if there is a factorization of $f$ as $$ X \overset{f'}\to Y' \overset{g} \to Y $$ in which $f'$ is an inclusion, $Y'$ is obtained from $X$ by iterated cell attachments of dimension $\le k$, and $g$ is a weak homotopy equivalence.

The notion of dimension I am describing is internal to the category of spaces under $X$, i.e, $X\backslash\text{Top}$, where $f$ is to be regarded as an object of that category. In this scheme, the homotopy dimension of the identity map is $\le -1$.

There is another variant of though: let define us say that the fiberwise homotopy dimension of $f: X\to Y$ is $\le k$ iff if there is a weak homotopy equivalence $X' \to X$ such that $X'$ is a cell complex of dimension $\le k$.

In particular $Y$ is homotopy equivalent to a cell complex of dimension $\le k$ if and only if the identity map of $Y$ has fiberwise dimension $\le k$.

Regarding Question 2: To a certain extent, I have written about both of these notions in the paper: Poincaré duality embeddings and fiberwise homotopy theory, Topology 38, 597$-$620 (1999), but this is by no means my concept, nor is my treatment to be regarded as definitive.

Added: The above notions generalize to a single notion as follows: let $f: A \to Y$ be any map of spaces and define $\text{Top}_f$ to be the category of factorizations of $f$ the objects of this category are factorizations $A \to X \to Y$ and morphisms are maps $X \to X'$ commuting with the given structure maps.

Then we can define dimension in this setting as follows: let's say that an object $X \in \text{Top}_f$ has dimension $\le k$ iff it is built up from the initial object (represented by $A$) by attaching cells over $Y$ of dimension at most $k$.

It's easy to see that the case $f:\emptyset \to Y$ gives the notion of fiberwise dimension, whereas the case when $f: A \to \text{pt}$ gives the notion of relative dimension.

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What about the homotopy dimension of the homotopy fiber of $f$ ? – BS. Feb 28 '11 at 15:30
That's not really a very useful notion, since the homotopy fiber is usually infinite dimensional. – John Klein Feb 28 '11 at 15:36
John, I edited to fix what I am sure were a couple of typos in your second-to-last paragraph. But how does $Y$ come into this? If $X$ admits an equivalence $X'\to X$ from a $k$-dimensional cell complex $X'$ then for any map $X\to Y$ we can also consider $X'$ as a space over $Y$. In other words the (fiberwise) homotopical dimension of $X$ as a space over $Y$ is the same as the homotopical dimension of $X$ as a space. – Tom Goodwillie Mar 1 '11 at 4:08
Yes, you're right---how silly of me. I'll fix that. – John Klein Mar 1 '11 at 4:34

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