Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The following result is sometimes known as the Compression Lemma:

Let $(X,A)$ be a CW pair and let $(Y,B)$ be a topological pair with $B\neq\emptyset$. For all $n$ such that $X-A$ has $n$-cells, assume that $\pi_n(Y,B,b_0)=0$ for all choices of basepoint $b_0\in B$. Then every map $f\colon\thinspace (X,A)\to (Y,B)$ is homotopic rel $A$ to a map with image in $B$.

See Hatcher's "Algebraic Topology", Section 4.1, or G.W. Whitehead's "Elements of Homotopy Theory", Section II.3. Using this result gives a quick route to J.H.C. Whitehead's theorem that a weak homotopy equivalence between CW complexes is a homotopy equivalence.

I am wondering whether there is a strengthening of this result which applies to individual maps, along the following lines:

Let $(X,A)$ be a CW pair with $A\neq \emptyset$ and let $(Y,B)$ be a topological pair. Let $f\colon\thinspace (X,A)\to (Y,B)$ be a map of pairs. For all $n$ such that $X-A$ has $n$-cells, assume that the induced map $f_\ast\colon\thinspace\pi_n(X,A,a_0)\to \pi_n(Y,B,f(a_0))$ is zero for all choices of basepoint $a_0\in A$, and [insert some extra conditions, possibly homological]. Then $f$ is homotopic rel $A$ to a map with image in $B$.

Does anyone know of such a result? Perhaps this is asking too much, and that the answer may well be "this is what obstruction theory does", but I thought I'd ask just in case.

share|improve this question
1  
I think this is just straightforward obstruction theory. If I'm right, this will be in Whitehead's account of obstruction theory. I'll check when I have the book and some time, if I remember to. –  Jeff Strom Feb 19 '13 at 13:57
    
If we convert the map $B \rightarrow Y$ to a fibration without changing the homotopy type, then this should follow from the discussion after Prop 4.74 in Hatcher (really that whole section.) Theorem 1 in section 4 of Spanier is also relevant... –  Dylan Wilson Feb 19 '13 at 14:14
1  
Some non-trivial extra conditions would certainly be required, as there exists a map $f:X\to Y$ between simply connected spaces which is not null-homotopic, yet induces the trivial homomorphism on all homotopy groups. The simplest example I know is the map $K({\mathbb Z},2)\to K({\mathbb Z},4)$ classifying the cohomology class $u^2\in H^4(K({\mathbb Z},2),{\mathbb Z})$ given by the square of a generator $u\in H^2(K({\mathbb Z},2),{\mathbb Z})\simeq {\mathbb Z}$. –  Ricardo Andrade Feb 20 '13 at 6:36

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.