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Let $ \ n \in \mathbb{N}$, $n \geqslant 2$. Equip $\mathbb{R}^n$ with the usual euclidean norm $ \ \Vert . \Vert : \mathbb{R}^n \to \mathbb{R}$, metric and topology. Consider $ \ p_0 = (1,0,0,...,0,0) \in \mathbb{R}^n \ $ and subspaces

$D^n = \big\{ x \in \mathbb{R}^n : \Vert x \Vert \leqslant 1 \big\} \qquad $ the unit disk,

$S^{n-1} = \big\{ x \in \mathbb{R}^n : \Vert x \Vert = 1 \big\} \qquad $ the unit sphere,

$S^{n-1}_+ = \big\{ (x_1,...,x_n) \in S^{n-1} : x_n \geqslant 0 \big\} \qquad $ the northern hemisphere,

$S^{n-1}_- = \big\{ (x_1,...,x_n) \in S^{n-1} : x_n \leqslant 0 \big\} \qquad $ the southern hemisphere and

$S^{n-2} = S^{n-1}_+ \cap S^{n-1}_- \qquad $ the equator.

Then, $p_0 \in S^{n-2}$, $S^{n-1}_+ \subset S^{n-1}$, $S^{n-1}_- \subset S^{n-1} \ $ and $ \ S^{n-1} = \partial D^n \subset D^n$.


Recall that a (basepointed) triad $ \ \textbf{X} =(X;A,B,x_0) \ $ consists of a topological space $X$, subspaces $A$ and $B$ and a basepoint $ \ x_0 \in A \cap B$. Note that $ \ \textbf{D}^n =(D^n;S^{n-1}_+,S^{n-1}_-,p_0) \ $ is a triad.

If $ \ \textbf{Y} =(Y;V,W,y_0) \ $ and $ \ \textbf{X} =(X;A,B,x_0) \ $ are triads, then a morphism of triads $ \ f : \textbf{Y} \to \textbf{X} \ $ is a continuous function $ \ f:Y \to X \ $ such that $ \ f[V] \subset A \ $, $f[W] \subset B \ $ and $ \ f(y_0) = x_0$. Denote by $ \ Hom \big( (Y;V,W,y_0) , (X;A,B,x_0) \big) = Hom( \textbf{Y}, \textbf{X}) \ $ the set of all morphisms of triads of the form $ \ \textbf{Y} =(Y;V,W,y_0) \to (X;A,B,x_0) = \textbf{X}$.

Let $ \ I= [0,1] \subset \mathbb{R} \ $ be the unit interval, equipped with the subspace topology of the usual euclidean topology of $\mathbb{R}$ and let $ \ f,g: (Y;V,W,y_0) \to (X;A,B,x_0) \ $ be morphisms of triads. Consider the topological space $ \ Y \times I \ $ with the product topology. We say that $f$ is triad-homotopic to $g$, and we write $ \ f \sim g$, if, and only if, there exists a homotopy $ \ H : Y \times I \to X \ $ such that, for each $ \ t \in I$, the continuous functions $ \ H_t : Y \to X \ $ such that $ \ H_t(y) = H(y,t)$, $\forall y \in Y$, satisfies $ \ H_0 = f$, $H_1 = g \ $ and $ \ H_t \in Hom \big( (Y;V,W,y_0) , (X;A,B,x_0) \big)$, $\forall t \in I$. The relation of triad-homotopy, $\sim$, is an equivalence relation.

Given a triad $ \ \textbf{X} =(X;A,B,x_0)$, define $$\pi_n \textbf{X} =\pi_n(X;A,B,x_0) = \pi_n(X;A,B) = \ Hom(\textbf{D}^n , \textbf{X}) \Big/ \! \! \! \sim$$ Recall that this quotient set is naturally a group for $ \ n\geqslant 3$, and is abelian for $ \ n \geqslant 4$. The group $ \ \pi_n \textbf{X} \ $ is called the "$n-$dimensional homotopy group of the triad $ \ \textbf{X} =(X;A,B,x_0)$".


Homotopy groups of triads are defined using $ \ \textbf{D}^n =(D^n;S^{n-1}_+,S^{n-1}_-,p_0)$. My question is:

How can one define the same notion based on a more general triad $ \ \textbf{Y} =(Y;V,W,y_0)$? That is, in some way that, for all $ \ n \geqslant 2$, we have isomorphisms $$\pi_n \textbf{X} \cong Hom(\textbf{Y}, \textbf{X}) \Big/ \! \! \! \sim$$ where $ \ Hom(\textbf{Y}, \textbf{X}) \Big/ \! \! \! \sim \ $ is equipped with some natural group product.

I thought maybe we need $Y$ homeomorphic to some compact and pathwise connected subset of $\mathbb{R}^n$, with $V$ and $W$ compact and pathwise connected subsets of $Y$ such that $S^{n-2}$ is a retract of the homeomorphic image of $ \ V \cap W \ $ in $D^n$, but I don't know if this will work and I don't know if this is the most general case. Can someone help me or show me some way how to do that?

Thanks in advance.

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  • $\begingroup$ I streamlined the exposition, your original post was a little cluttered. I hope you agree with the changes. You can always edit the post to your liking. $\endgroup$ – Olivier Bégassat Sep 20 '14 at 6:03
  • $\begingroup$ Possibly depending on how good the spaces are, if your new triad is triad homotopic to $\mathcal{D}^n$ you will get the space result. $\endgroup$ – Najib Idrissi Sep 20 '14 at 6:45
  • $\begingroup$ Thanks Olivier Bégassat, I merged the two texts because I did not liked your style, but I used some parts of your writings. $\endgroup$ – Gustavo Sep 20 '14 at 23:13
  • $\begingroup$ The space $Y$ need to be good enough for us to equip $ \ Hom ( \textbf {Y}, \textbf {X} ) \ $ with a group product for $ \ n \geqslant 3 \ $ and that this product is commutative for $ \ n \geqslant 4$. $\endgroup$ – Gustavo Sep 20 '14 at 23:20

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