# Suspension homomorphism

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.

• It is not clear what you want to ask. Is it about the definition of the suspension homomorphism? Jul 14, 2021 at 8:44
• @PaulFrost Edited the question, hope is clearer. Jul 14, 2021 at 8:47
• Typo: $\pi_{i+1}(CX,X,\star)$ Jul 14, 2021 at 10:19
• @PaulFrost Edited the typo, thanks Jul 14, 2021 at 16:10
• @PaulFrost My problem here really concerns commutativity. I was able to prove 1 and 2. I don't see see how commutativity holds since $\partial$ is not explicit here. Am I right? Jul 14, 2021 at 22:08

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 $$(Y,y_0) = (S^i,*)$$).

• I'm sorry to bother you, what I don't get is what the overline in the last sentence means and how you are able to verify the last equality, element by element? i.e for every $g \in \pi_i(X)?$ Jul 15, 2021 at 15:59
• @jacopoburelli I associated to each $\phi : (D^{i+1},S^i,*) \to (C(X,x_0)/X,*,*)$ a map $\bar \phi : (S^{i+1},*) = (D^{i+1}/S^i,*) \to (C(X,x_0)/X,*)$. This is possible because $\phi(S^i) = *$. Apply this to $q \circ Cg$ to get $\overline{q \circ Cg}$. Can you see why this map is the same as $\Sigma(g)$? Jul 15, 2021 at 16:05
• @jacopoburelli I made an edit. Jul 15, 2021 at 16:30
• Thanks, I'm going to think about it, your answers are explanatory as always Jul 15, 2021 at 16:43
• I was not able to verify that $\overline{q \circ C_g}$ is the same as $\Sigma(g)$ because there a lot of identifications between cones and suspension. How should I proceed? Aug 11, 2021 at 8:28