7
$\begingroup$

The notion of a relative Whitehead product exists in the literature and has been asked about before (e.g. here). I am trying to find out about a different product on relative homotopy classes which could also be referred to as a relative Whitehead product, but which I cannot find in the literature.

Let $X$ be a CW complex with $A \subseteq X$ a subcomplex. Let $f:(D^n,S^{n-1}) \rightarrow (X,A)$ and $g:(D^m,S^{m-1}) \rightarrow (X,A)$ be representatives of relative homotopy classes $[f] \in \pi_n(X,A)$ and $[g] \in \pi_m(X,A)$, for $m,n \geq 2$.

By writing $(D^n,S^{n-1}) = (\Sigma D^{n-1},\Sigma S^{n-2})$ and $(D^m,S^{m-1}) = (\Sigma D^{m-1},\Sigma S^{m-2})$ we obtain the adjoint maps of pairs $\hat{f}: (D^{n-1},S^{n-2}) \rightarrow (\Omega X, \Omega A)$ and $\hat{g}:( D^{m-1}, S^{m-2}) \rightarrow (\Omega X, \Omega A)$. By using the loop multiplication and then taking the adjoint back, we obtain a representative $(\hat{f} \wedge \hat{g})^{\wedge}: (D^{m+n-1},S^{m+n-2}) \rightarrow (X,A)$ of the relative homotopy class $[(\hat{f} \wedge \hat{g})^{\wedge}] \in \pi_{m+n-1}(X,A).$

This is essentially a direct analogue of one of the ways in which we can construct the usual two-fold Whitehead product.

I can't find anything in the literature about ``relative" Whitehead products like this. Everything I've seen (e.g. in Golasinski, de Melo) seems to address only so-called "mixed" relative Whitehead products, where one of $[f]$ and $[g]$ is a relative class, and the other a regular homotopy class.

$\endgroup$
4
  • 4
    $\begingroup$ The relative homotopy groups are just the (degree shifted :() homotopy groups of the homotopy fiber, you are simply describing the Whitehead product there. $\endgroup$ Commented Mar 24, 2021 at 13:47
  • $\begingroup$ I see. Can I just check that we are speaking of the homotopy fiber of the inclusion $A \rightarrow X$, and then this follows from considering the long exact sequence of the fibration? $\endgroup$
    – Matt
    Commented Mar 25, 2021 at 13:14
  • $\begingroup$ Yes, that's precisely it (but you can also prove it directly from the definitions, it's a not too hard exercise often given to the students) $\endgroup$ Commented Mar 25, 2021 at 13:22
  • 2
    $\begingroup$ Wouldn't the Whitehead product on the homotopy fiber land in $\pi_{n+m-2}(X,A)$? $\endgroup$ Commented Mar 26, 2021 at 17:54

0

You must log in to answer this question.