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It is known (see for instance https://math.stackexchange.com/a/42020) that "the homotopy type of a CW complex is entirely determined by the homotopy classes of the attaching maps". What is the precise statement of this result? Can it be found somewhere?

Related to this, is there a (more or less "established") notion of isomorphism between CW-complexes? Something like the following:

  • $X \sim Y$ if $Y$ is obtained from $X$ changing the attaching map of some cell $e^n$ by homotopy (and somehow adapting the attaching maps of higher-dimensional cells);

  • $X$ is isomorphic to $Y$ if $X=X_1 \sim X_2 \sim \dots \sim X_n=Y$.

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Giove
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  • The resulting complexes will have the same homotopy type. I think that topologists use CW complexes for some kind of combinatorial description of homotopy types of spaces. "Having the same homotopy type" is, however, a much more general equivalence relation then what you propose. – Peter Franek Apr 06 '15 at 10:52

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The key to this statement can be found, for example, in Hatcher's book Algebraic Topology:

Proposition 0.18. If $(X_1, A)$ is a CW pair and we have attaching maps $f,g : A \to X_0$ that are homotopic, then $X_0 \cup_f X_1 \simeq X_1 \cup_g X_0 \, \mathrm{rel} \, A$.

Here "CW pair" means that $X_1$ is a CW complex and $A$ is a subcomplex. Now you can apply this proposition to $(X_1, A) = (D^n, \partial D^n) = (D^n, S^{n-1})$ to see that if an $n$-cell is attached to a space $X_0$ by two homotopic attaching maps $f,g : S^{n-1} \to X_0$, then the resulting spaces $X_0 \cup_f e^n$ and $X_0 \cup_g e^n$ are homotopy equivalent.

Now it is simply a matter of induction and bookkeeping (and if your CW complex is not finite, you need to recall the definition of the weak topology; I'll focus on the finite case here). In small dimensions:

  • Start with the $0$-skeleton $X_{(0)}$.
  • Maps $S^0 \to X_{(0)}$ are homotopic iff they are equal, so WLOG you can actually start with the $1$-skeleton $X_{(1)}$.
  • Now suppose you attach two-cells via maps $f_\alpha : S^1 \to X_{(1)}$ to get $X_{(2)}$, and that you also have maps $g_\alpha : S^1 \to X_{(1)}$ to get $X_{(2)}'$. Suppose that $f_\alpha \sim g_\alpha$ for all $\alpha$; then by the proposition $X_{(2)} \simeq X_{(2)}'$.
  • Now you attach three-cells via maps $f_\beta : S^2 \to X_{(2)}$ and $g_\beta : S^2 \to X_{(2)}'$ to respectively get $X_{(3)}$ and $X_{(3)}'$. Since $X_{(2)} \simeq X_{(2)}'$, it follows that $[S^2, X_{(2)}] \cong [S^2, X_{(2)}']$ (where $[X,Y]$ is the set of homotopy classes of maps). If in this set $[f_\beta] = [g_\beta]$ for all $\beta$, it again follows from the proposition that $X_{(3)} \simeq X_{(3)}'$.

You can continue this reasoning by induction: a CW complex is uniquely determined by its $1$-skeleton, by homotopy classes of attaching maps in $[S^1, X_{(1)}]$, then by homotopy classes of attaching maps $[S^2, X_{(2)}]$, etc. This is what is meant by the statement.

Note that the CW complexes obtained are not isomorphic (I guess that would mean a cellular homeomorphism), they are homotopy equivalent, in the sense that they have the same homotopy type.

Najib Idrissi
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