Suspension homomorphism

Question

Let $(X,x_0)$ be a pointed space. Consider the suspension map $\Sigma : \pi_i(X,x_0) \longmapsto \pi_{i+1}(SX,x_0)$, where $SX \simeq X \times I/(X \times \partial I \cup x_0 \cup I)$.

In order to prove that the suspension map is an homomorphism we need the following :

1. The topological cone $CX$ is contractible
2. $CX/X \simeq SX$
3. The following diagramm commutes :

$\require{AMScd}$ \begin{CD} 0=\pi(CX,\star) \\ @AAA\\ \pi_i(X,\star)@>{\Sigma}>> \pi_{i+1}(SX,\star) \\ @AA\partial A @VVV \\ \pi_{i+1}(CX,X,\star) @>{q_{*}}>> \pi_{i+1}(CX/X,\star) \\ @AAA \\ 0=\pi_{i+1}(CX,\star) \end{CD}

My question concerns understanding those three points better.

It thinks the idea behind 1. is to deformation retract onto the base point, but I think the question has already an answer here the cone is contractible, so this point is okay.

I don't understand if the commutativity of the diagram follows from some naturality of long exact sequences in homotopy. I recognise the sequence of the pair $(X,CX)$ on the vertical left arrow but I don't see how to fit the vertical arrow on the right properly in order to have a commutative diagram.

Any help or hint on this and also of point 2. (which I don't even visualize) would be appreciated.

Answer 1

Concerning 1. : Note that in the context of your question $CX$ denote the reduced cone $$C(X,x_0) = X \times I/(X \times \{1\} \cup \{x_0\} \times I) .$$ In the linked question $CX$ denotes the unreduced cone $CX = X \times I/X \times \{1\}$. Note that $X$ can be identified with $p(X \times \{0\}) \subset C(X,x_0)$, where $p : X \times I \to C(x,x_0)$ is the quotient map. Laxly speaking, $X$ is identified with the base of the reduced cone. With respect to this identification $x_0 \in X$ is identified with $* = [X \times \{1\} \cup \{x_0\} \times I] \in C(x,x_0)$ which is the tip of the reduced cone.

Anyway, also the reduced cone is contractible to $* \in C(X,x_0)$ (the proof for the unreduced case easily transfers to the reduced case).

Concerning 2. : Yes, one can even take $S(X,x_0) = C(X,x_0)/X$. Let $q^{(X,x_0)} : C(X,x_0) \to S(X,x_0)$ denote the quotient map.

Clearly a pointed map $h : (X,x_0) \to (Y,y_0)$ induces a pointed map $Ch : (C(X,x_0),*) \to (C(Y,y_0),*)$. This map is an extension of $h : X \to Y$ where we regard $X$ as the base of $C(X,x_0)$ and $Y$ as the base of $C(Y,y_0)$. Hence $Ch$ induces a unique map $\Sigma h : S(X,x_0) \to S(Y,y_0)$; it is charactertized by the property $\Sigma h \circ q^{(Y,y_0)} = q^{(X,x_0)} \circ C h$.

Now let us prove 3. which shows that $\Sigma$ is a homomorphism. Note that the right vertical arrow can be replaced by $=$ by our remark concerning 2.

We know that $\partial : \pi_{i+1}(C(X,x_0),X,*) \to \pi_i(X,x_0)$ is an isomorphism. The elements of $\pi_{i+1}(C(X,x_0),X,*)$ are homotopy classes of maps of triples $f : (D^{i+1},S^i,*) \to (C(X,x_0),X,*)$ and $\partial$ is given by restriction, i.e. $\partial ([f]) = [f : (S^i,*) \to (X,x_0)]$. Therefore, since $\partial$ is an isomorphism in the present case, we can easily compute $\partial^{-1}([g])$ for $[g] \in \pi_i(X,x_0)$ as follows:

The map $Cg : (D^{i+1},*) = C(S^i,*) \to C(X,x_0)$ has the property $Cg(S^i) \subset X$; in fact, on $S^i \subset C(S^i,*)$ it agrees with $g$. It can therefore be regarded as a map of triples $Cg : (D^{i+1},S^i,*) \to (C(X,x_0),X,*)$. By construction $\partial([Cg]) = [g]$.

Now consider the quotient map $q^{(X,x_0)} : C(X,x_0) \to C(X,x_0)/X = S(X,x_0)$. It induces $q^{(X,x_0)}_* : \pi_{i+1}(C(X,x_0),X;*) \to \pi_{i+1}(C(X,x_0)/X,*,*)$. The latter group can be identified with $\pi_{i+1}(C(X,x_0),*)$ because maps of triples $\phi : (D^{i+1},S^i,*) \to (C(X,x_0)/X,*,*)$ can be identified with maps of pairs $\bar \phi : (S^{i+1},*) = (D^{i+1}/S^i,*) \to (C(X,x_0)/X,*)$. In fact, since $\phi(S^i) = *$, it induces a unique $\bar \phi : (D^{i+1}/S^i,*) \to (C(X,x_0)/X,*)$ characterized by the property $\bar \phi \circ q^{(S^i,*)} = \phi$. A similar identificaton works for homotopies.

But now by construction $q^{(X,x_0)}_*([Cg]) = [q^{(X,x_0)} \circ Cg]$. Moreover, you can easily see that $\overline{q \circ Cg} = \Sigma g$ because $\overline{q^{(X,x_0)} \circ Cg}$ is characterized by the property $\overline{q \circ Cg} \circ q^{(S^i,*)} = q^{(X,x_0)} \circ Cg$.

Update:

We have $$S(X,x_0) = X \times I/(X \times \{0, 1\} \cup \{x_0\} \times I) .$$ Given a basepoint-preserving map $g : (Y,y_0) \to (X,x_0)$ the map $g \times id_I : Y \times I \to X \times I$ induces basepoint-preserving maps $$Cg : C(Y,y_0) \to C(X,x_0) ,$$ $$Sg : S(Y,y_0) \to S(X,x_0)$$ because the subspaces of $Y \times I$ which are collapsed to points in the quotients $C(Y,y_0)$ resp. $S(Y,y_0)$ are mapped by $g \times id_I$ to the corresponding subspaces of $X \times I$.

As above we can identify $S(Y,y_0)$ with $C(Y,y_0)/Y$ and $S(X,x_0)$ with $C(X,x_0)/X$. Note that $Cg$ maps the base $Y$ of $C(Y,y_0)$ into the base $X$ of $C(X,x_0)$. It is then clear from the definition of $Cf$ and $Sf$ that the following diagram commutes:

$\require{AMScd}$ \begin{CD} C(Y,y_0) @>{Cg}>> C(X,x_0) \\ @V{q^{(Y,y_0)}}VV @V{q^{(X,x_0)}}VV \\ S(Y,y_0) @>{Sg}>> S(X,x_0) \end{CD}

This means that $Sg$ is the map I denoted by $\overline{q^{(X,x_0)} \circ Cg}$ (in the special case $(Y,y_0) = (S^i,*)$).