# Homotopy dimension of a mapping

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 $h\dim 1_X$ $<$$\infty while h\dim X=\infty. Question 2: I am sure that this is a well-known and well-studied notion, and that I am merely using the wrong search terms. Where should I look in the literature to learn more about this concept? - ## 1 Answer Regarding Question 1: No, I do not think that's correct. In my opinion, the definition should be one of the following: The relative homotopy dimension of f: X \to Y is \le k if and only if there is a factorization of f as$$ 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}$gives the notion of relative dimension. - What about the homotopy dimension of the homotopy fiber of$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