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Working in the context of James Construction, using the following lemma:

If $X$ and $Y$ and connected CW-complexes then $$S(X \times Y) \simeq SX \vee SY \vee S(X \wedge Y)$$

and the diagram $\begin{matrix} FW_k(X)&\to &X^k &\to X^{(k)}\\\downarrow &&\downarrow & \downarrow\\ J_{K-1}(X) &\to&J_k(X) &\to X^{(K)}\end{matrix} $

where $FW_k(X)$ is the $k$th fat wedge of $X$ and $X^{(K)} = J_k(X) / J_{K-1}(X)$, I don't understand the proof of the following proposition :

Proposition: If $X$ has the homotopy type of a connected CW-complex then $$ S(J(X)) \simeq \bigvee_{k=1}^{\infty} S(X^{(k)})$$

Proof: Using the previous lemma and induction on $k$ the top right epimorphism in the preceding diagram has a homotopy splitting after suspendingm and so the bottom does as well.

I really dont understand this proof, most likely because I don't understand what the author means by "homotopy splitting".

Note : this comes from Paul Selick, Introduction to Homotopy Theory.

Oscar P.
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  • By "homotopy splitting" he means in this context a right homotopy inverse. That is, a map $\varphi:X^{(k)}\rightarrow X^k$ such that $q\circ \varphi\simeq id_{X^{(k)}}$, where $q:X^k\rightarrow X^{(k)}=\wedge^k X$ is the quotient. – Tyrone May 07 '18 at 20:51
  • Thanks for your answer, but after some more thought I think he means by "homotopy splitting" that you can write the space has a wedge of other spaces with the samehomotopy type. That is you split your space into smaller ones. – Oscar P. May 09 '18 at 09:10
  • The wording of the question suggests the he is interested in the map rather than the space, and a homotopy splitting of a map is the definition I have given above. – Tyrone May 09 '18 at 09:54
  • In this case they are the same, however. The homotopy equivalence $f:\Sigma(X\times Y)\xrightarrow{\simeq} \Sigma X\vee \Sigma Y\vee \Sigma(X\wedge Y)$ is obtained by adding together the three maps $\Sigma pr_X$, $\Sigma pr_Y$, and $\Sigma q$ using the co-H-structure given by the suspension on $\Sigma(X\times Y)$. Hence the right homotopy inverse of $\Sigma q$ may be obtained as the composite $\Sigma (X\wedge Y)\hookrightarrow \Sigma X\vee\Sigma Y\Sigma (X\wedge Y)\xrightarrow{f^{-1}}\Sigma(X\times Y)$. – Tyrone May 09 '18 at 09:54
  • It is not always the case, however, that a homotopy splitting is induced from a wedge splitting of the space. Consider for example the projection $pr_X:X\times Y\rightarrow X$. For for a more intricate example, the map induced by the determinant $Bdet:BU(2)\rightarrow BS^1$. There is no decomposition of $BU(2)$ as either a product or a wedge of any other non-trivial spaces, although this map has a right homotopy inverse. – Tyrone May 09 '18 at 09:55
  • @Tyrone How does the co-H-structure give the equivalence? I haven't seen this argument before, do you have a reference? – Ali Caglayan Sep 10 '19 at 12:59
  • @AliCaglayan it doesn't give the equivalence, it only produces the map $f$. The upside is that constructed in this way you know exactly what $f$ does on homology. Now just use the Whitehead theorem for homology, since if $X,Y$ are connected, then both the domain and codomain of $f$ are simply connected. – Tyrone Sep 10 '19 at 13:34

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