MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I wanted to ask if there is any known mehod to quantify 'how many' homotopy classes of maps there are between two given topological spaces $X$, $Y$ (CW-complexes, say). So far I had the following idea: If two maps $f,g:X\rightarrow Y$ are homotopic, then they induce the same maps on all homotopy groups. Maybe there's a class of spaces for which the converse is also true. Then I would have a characterization via Homomorphisms between homotopy groups.

Does anybody know if there is a class of spaces for which this is true?

share|cite|improve this question
I voted to close, because it is wellknown: – Marc Palm Jul 18 '13 at 13:49
No this isn't Withehead's theorem. Whitehead's theorem says that if a map induces an isomorphism on all homotopy groups then it is a homotopy equivalence. – quiver Jul 18 '13 at 14:15
Well, I thought Marc meant it can be deduced from Whitehead's theorem. It seems to me one might consider the pullback $P$ of $\langle f, g \rangle: X \to Y \times Y$ along the path fibration $Y^I \to Y \times Y$ and show that the projection $P \to X$ induces an isomorphism on all homotopy groups, and then apply Whitehead. – Todd Trimble Jul 18 '13 at 14:35
@Todd: the homotopy fibre of $P \to X$ is equivalent to the loop space $\Omega Y$, so $P \to X$ will rarely induce an isomorphism on homotopy groups. – Oscar Randal-Williams Jul 18 '13 at 15:08
An old question of mine seems relevant: – David Speyer Jul 18 '13 at 15:56

How about Eilenberg--Mac Lane spaces?

Let $G$ and $H$ be any groups. For pointed homotopy classes, $\langle K(G,1), K(H,1)\rangle $ is $\operatorname{Hom}(G,H)$, and for unpointed homotopy classes $[K(G,1), K(H,1)]$ is $\operatorname{Hom}(G,H)/H$, the orbits under conjugation by $H$.

When $n>1$ and $G$ and $H$ are abelian, we have $$\langle K(G,n),K(H,n)\rangle = [K(G,n), K(H,n)]=\operatorname{Hom}(G,H),$$ as pointed out by András in the comments.

Eilenberg--Mac Lane spaces also give counter-examples to the conjecture that maps $f,g:X\to Y$ between CW complexes are homotopic if they induce the same maps on homotopy groups. See my answer here.

share|cite|improve this answer
I guess you don't even need the source to be an Eilenberg-Mac Lane space for this. $\langle X,K(G,1) \rangle$ is $\mathrm{Hom}(\pi_1(X),G)$. – Dan Petersen Jul 18 '13 at 17:22
I think 1 can be replaced by any $n.$ I mean that $[K(G,n), K(H,n)] = Hom (G,H)$ – András Szűcs Jul 18 '13 at 20:13
Good comments. I have edited the answer to include $n>1$. (Welcome to MO, András!) – Mark Grant Jul 19 '13 at 9:53

Vidit's comments are relevant to a paper of Graham Ellis

Homotopy classification the J.H.C. Whitehead way. Exposition. Math. {6} (1988) 97--110.

The more general result given here is that if $C$ is a crossed complex and $X$ is a CW-complex then there is a bijection of homotopy classes

$$[X,BC] \cong [\Pi X_*, C]$$

where $BC$ is the classifying space of the crossed complex, and $\Pi X_*$ is the fundamental crossed complex of the skeletal filtration of $X$. A cubical version of the proof is given in the EMS Tract Vol 15 Nonabelian Algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids, where a pdf is available. This book also contains our generalisation of the work in CHII on the relation between crossed complexes and chain complexes with a group of operators. He writes in essence that the former have better realisation properties, and the latter are better for calculation.

Note that the construction $BC$ generalises Eilenberg-Mac Lane spaces, including the local system case.

This has been generalised to the equivariant case,

Brown, R., Golasinski, M., Porter, T. and Tonks, A. Spaces of maps into classifying spaces for equivariant crossed complexes. II. The general topological group case. $K$-Theory 23 (2001) 129--155. arxiv 9808111

There is a lot in Whitehead's CHII ! One example is his result on free crossed modules. I wrote up his proof ([30] in my publication list) and the referee wrote that: "The theorem is not new, the proof is not new, and the paper should be published." This theorem is sometimes stated in texts on algebraic topology but rarely proved. Note that Whitehead's "homotopy systems" are our "reduced free crossed complexes".

To go back to the original question, it must be said that the homotopy groups are but a pale shadow of the homotopy type, and indeed crossed complexes give only a "linear" perspective, though linear methods are often useful as an approximation.

The fist displayed equivalence is part of a weak equivalence

$$ B(CRS(\Pi X_*, C )) \to (BC) ^X$$

which does give information on homotopies in this case. See the above mentioned book, Theorem 11.4.19.

share|cite|improve this answer
Ronnie, is there an online version of Ellis' article? – Vidit Nanda Jul 19 '13 at 14:37
I have one available for asking. – Ronnie Brown Jul 26 '13 at 19:59

Mark Grant's answer provides a good class of examples, but a slightly more general class can be found after (many, many hours) of reading Whitehead's "Combinatorial Homotopy II" available here.

I think the content of the Corollary right after Theorem 6 (on page 468 of the linked document) is as follows after one unzips the notation:

If $X$ and $Y$ are connected CW complexes with $\dim X$ smaller than $n$ and all homotopy groups $\pi_kY$ trivial for $k$ between $2$ and $n-1$, then the homotopy classes of maps $[X,Y]$ correspond bijectively to $\pi_1Y$-equivariant chain homotopy classes $[C(X'),C(Y')]_{\pi_1Y}$ where $X'$ and $Y'$ are the universal covers of $X$ and $Y$ respectively. From this point onwards, the problem reduces to linear algebra (although I'm not claiming that this algebra is algorithmically tractable in general).

I hope someone who can actually read Whiteheadese will confirm that I have not screwed this up.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.