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Let $X$ be a Hausdorff space and $A$ a closed subspace. Suppose the inclusion $A \hookrightarrow X$ is a cofibration. Let $f, g: X \to Y$ be maps that agree on $A$ and which are homotopic. Are they homotopic relative to $A$?

My motivation for asking this question comes from the following result:

Let $i: A \to X, j: A \to Y$ be cofibrations. Suppose $f: X \to Y$ is a map which makes the natural triangle commutative. Suppose $f$ is a homotopy equivalence. Then $f$ is a cofiber homotopy equivalence.

On the other hand, I'm having trouble adapting the proof in Peter May's book of this to the question I asked. Nonetheless, the standard examples of pairs of maps which are homotopic but not with respect to which some subset on which they agree (say, the identity map of a comb space and its collapsing to a suitably chosen point), don't seem to involve NDR pairs.

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1 Answer 1

up vote 5 down vote accepted

I think the machinery of obstruction theory deals with the special case where the spaces are skeleta of a CW-complex. Here's the setup (I'm basically just copying from pp. 6-7 of Mosher & Tangora here):

Let $Y$ be simply-connected for simplicity. First, let $B$ be a complex and $A$ be a subcomplex. Let $f:A\cup B^n\rightarrow Y$. Then we get an obstruction cochain $c(f)\in C^{n+1}(B,A;\pi_n(Y))$ (i.e. a function on relative $(n+1)$-cells with values in $\pi_n(Y)$). Similarly, let $K$ be a complex; then for any two maps $f,g:K\rightarrow Y$ that agree on $K^{n-1}$, we similarly get a difference cochain $d(f,g)\in C^n(K;\pi_n(Y))$.

Here are the two results.

  • Theorem: There is a map $g:A\cup B^{n+1}\rightarrow Y$ agreeing with $f$ on $A\cup B^{n-1}$ iff $[c(f)]=0 \in H^{n+1}(B,A;\pi_n(Y))$.
  • Theorem: The restrictions of $f$ and $g$ to $X^n$ are homotopic rel $K^{n-1}$ iff $d(f,g)=0 \in C^n(K;\pi_n(Y))$. They are homotopic rel $K^{n-2}$ iff $[d(f,g)]=0 \in H^n(K;\pi_n(Y))$.
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I just realized that this doesn't explicitly address your question -- the first theorem is about extending the definition of a map, and the second is about two maps that may or may not be homotopic (although probably you could use this to create a counterexample). Here's an easy one. Let $X=[0,1]$, $A=\{0,1\}$, $Y=S^1$. Then the degree $\pm 1$ maps $(X,A)\rightarrow (Y,*)$ agree on $A$, are homotopic, but are not homotopic rel $A$. –  Aaron Mazel-Gee Oct 10 '10 at 19:32
Sorry for the slow response: thanks! That's a nice example. –  Akhil Mathew Oct 11 '10 at 13:03

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