# Definition of the triad homotopy groups

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.

• 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. – Olivier Bégassat Sep 20 '14 at 6:03
• 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. – Najib Idrissi Sep 20 '14 at 6:45
• Thanks Olivier Bégassat, I merged the two texts because I did not liked your style, but I used some parts of your writings. – Gustavo Sep 20 '14 at 23:13
• 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 $\ n \geqslant 4$. – Gustavo Sep 20 '14 at 23:20