Homotopy groups of a (pointed) topological space can be defined in multiple ways. In particular, I'm interested in proving rigourosly that the following two definitions are equivalent:
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$\pi_n(X,x_0)=[(\mathbf{I}^n,\partial\mathbf{I}^n),(X,x_0)]$ and the concatenation of two continous maps $f,g:(\mathbf{I}^n,\partial\mathbf{I}^n) \to (X,x_0)$ is:
$$f+g: \mathbf{I}^n \to X, \ \ \ \ (s_1,\ldots,s_n)\mapsto \begin{cases}f(2s_1,s_2,\ldots,s_n) \ \ \ \ \ \ \ \ \ \ \ \text{if $s_1\leq \frac12$} \\ g(2s_1-1,s_2,\ldots,s_n) \ \ \ \ \text{if $s_1\geq \frac12$} \end{cases}.$$ -
$\pi_n(X,x_0)=[(S^n,\ast),(X,x_0)]$ and the concatenation of two continous maps $\hat{f},\hat{g}:(S^n,\ast)\to (X,x_0)$ is:
$$\hat{f}+\hat{g}:=(\hat{f}\vee \hat{g})\circ \mu$$
Where $\hat{f} \vee \hat{g}:(S^n\vee S^n,\ast) \to (X,x_0) $ is the continous map, whom existence is assured by the universal property of the wedge sum and $\mu:S^n\to S^n\vee S^n$ is the projection that collapses a whole equator that contains $\ast$.
I managed to fill in some details. For example I managed to prove that the following map is a well defined bijection:
$$\mathcal{J}: [(\mathbf{I}^n,\partial\mathbf{I}^n),(X,x_0)] \to [(S^n,\ast),(X,x_0)], \ \ \ \ [f]\mapsto [\hat{f}] $$
Where $\hat{f}$ is the only continous map that makes the following diagram commutes:
$\hspace{4.5cm}$
Now I just need to prove that $\mathcal{J}[f+g]=[\hat{f}+\hat{g}]$, so I need a (pointed) homotopy between $\mathcal{J}[f+g]$ and $(f\vee g)\circ \mu$, but I'm having some trouble in constructing it. Could you help me?
Best Answer
It seems to be geometrically clear what the "pinching map" $\mu : S^n \to S^n \vee S^n$ looks like, but we have to be very precise.
Let us assume that the basepoint $*$ of $S^n$ lies in the equatorial sphere $S^{n-1}_0 = S^{n-1} \times \{0\} \subset S^n$. We may simply take $* = (1,0,\dots,0)$. Then the quotient $S^n /S^{n-1}_0 $ is certainly homeomorphic to the wedge of two copies of $S^n$, but this does not specify a concrete $\mu$.
Let us first construct a homeomorphism $D^n/S^{n-1} \to S^n$. The construction is technically non-trivial; it is a slight modification of that presented in my answer to Two topology questions regarding quotient $D^n/S^{n-1}$ and homotopy $S^{n-1} \to S^{n} - \{ a,b\}$. We write $S^n = \{ (r,x) \in \mathbb{R} \times \mathbb{R}^n \mid r^2 + \lVert x \rVert^2 = 1 \}$, where $\lVert - \rVert$ denotes the Euclidean norm.
Define $$p : D^n \to S^n, p(x) = \begin{cases} \left(2\lVert x \rVert -1, \dfrac{\sqrt{1 - (2\lVert x \rVert -1)^2}}{\lVert x \rVert}x \right) & x \ne 0 \\ (-1,0) & x = 0 \end{cases} $$ This is well-defined because $(2\lVert x \rVert -1) \in [-1,1]$ and $$ (2\lVert x \rVert-1)^2 + \left\lVert \dfrac{\sqrt{1 - (2\lVert x \rVert -1)^2}}{\lVert x \rVert}x \right\rVert^2= (2\lVert x \rVert -1)^2 + 1 - (2\lVert x \rVert -1)^2 = 1 .$$ $p$ is continuous because for $x \to 0$ we get $\left\lVert \dfrac{\sqrt{1 - (2\lVert x \rVert -1)^2}}{\lVert x \rVert}x \right\rVert = \sqrt{1 - (2\lVert x \rVert -1)^2} \to 0$ and $2\lVert x \rVert -1 \to -1$.
For $x \in S^{n-1}$ we have $p(x) = (1,0) = *$.
Next define $$j : S^n \setminus \{*\} \to D^n \setminus S^{n-1}, j(r,y) = \begin{cases} \dfrac{1 + r}{2\sqrt{1- r^2}}y & y \ne (-1,0) \\ 0 & y = (-1,0) \end{cases} $$ This is well-defined because $-1 < r < 1$ for $(r,y) \in S^n \setminus \{*, (-1,0)\}$ and $$\left\lVert \dfrac{1 + r}{2\sqrt{1- r^2}}y \right\rVert = \dfrac{1 + r}{2\sqrt{1- r^2}}\left\lVert y \right\rVert = \dfrac{1 + r}{2} < 1 .$$ It is easily verified that $p(j(r,y)) = (r,y)$ for all $(r,y)$ and $j(p(x)) = x$ for all $x \in D^n \setminus S^{n-1}$. Hence $p$ maps $D^n \setminus S^{n-1}$ bijectively onto $S^n \setminus \{*\}$.
$D^n , S^n$ are compact Hausdorff, hence $p$ is a closed map and thus a quotient map. By the above considerations there exists a unique function $h : D^n/S^{n-1} \to S^n$ such that $h \circ \pi = p$, where $\pi : D^n \to D^n/S^{n-1}$ is the standard quotient map. By construction it is a bijection. By the universal property of quotient maps $p$ and $\pi$ both $h$ and $h^{-1}$ are continuous, i.e. $h$ is a homeomorphism.
Geometrically $h$ can be visualized as follows: Each sphere $\Sigma^{n-1}_r = \{ x \in D^n \mid \lVert x \rVert = r \} \subset D^n$ with $0 < r < 1$ is mapped to the sphere $S^{n-1}_r = S^n \cap (\{2r-1\} \times \mathbb R^n) \subset S^n$, $S^{n-1}$ goes to $*$ and $0$ to $(-1,0)$. That is, $D^n$ is slipped like a rubber glove over $S^n$ beginning from the left (first coordinate $-1$) to the right (first coordinate $1$).
The equatorial ball $D^{n-1}_0 = D^{n-1} \times \{0\} \subset D^n$ is mapped by $h$ onto the equatorial sphere $S^{n-1}_0$. The other level balls $D^{n-1}_t = D^n \cap (\mathbb R^{n-1} \times \{r\})$ are mapped to "inclined subspheres" of $S^n$ which all meet in $*$.
Let us next observe that in definition 1 we can replace $I$ by $J = [-1,1]$ and obviously obtain an isomorphic group, i.e. we may take $$\pi_n(X,x_0) = [(J^n,\partial J^n),(X,x_0)]$$ with addition defined by $ [f] +[g)] = [f+g]$, where $$f+g: J^n \to X, (f+g)(x_1,\ldots,x_n) = \begin{cases}f(2x_1+1,x_2,\ldots,x_n) & x_1 \le 0 \\ g(2x_1-1,x_2,\ldots,x_n) & x_1 \ge 0 \end{cases}$$
There is an obvious homeomorphism $\phi : J^n \to D^n$; see my answer to $(D^n\times I,D^n \times 0)$ and $(D^n \times I, D^n \times 0 \cup \partial D^n \times I)$ are homeomorphic (note that $J^n$ and $D^n$ are the closed unit balls with respect to the maximum-norm $\lVert - \rVert_\infty$ and the Euclidean norm). It maps $J^{n-1}_0 = J^{n-1}\times \{0\}$ to $D^{n-1}_0$.
Consider the quotient map $q = p \circ \phi : J^n \to S^n$. Split $J^n$ into the two cuboids $J^n_\pm = \{(x_1,\ldots, x_n) \mid \pm x_1 \ge 0\}$. Then $q$ induces quotient maps $q_\pm : J^n_\pm \to S^n, q_\pm(x_1,\ldots, x_n) = q(2x_1 \mp 1, x_2,\ldots, x_n)$. Since they agree on $J^{n-1}_0$, they can be pasted to a map $Q : J^n \to S^n \vee S^n$. By construction $Q(\partial J^n) = *$, thus $Q$ induces $\mu : S^n \to S^n \vee S^n$. This is the desired explicit pinching map. By construction $\mu(S^{n-1}_0) = \{*\}$.
It is now a routine exercise to verify that your bijection $\mathcal J$ is a group homomorphism.