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I am trying to see whether it is possible to define smash product of infinite loop spaces using the space $S^{\infty}$.

Let C be the category of compactly generated weak Hausdorff topological spaces. For such a space X, define $\Omega^{\infty}X$ to be the space of based continuous maps out of $S^{\infty}$ into $X$. Being an infinite loop space equivalent to the existence of a space X such that X $\cong$ $\Omega^{\infty} X_0$

Suppose we define, for infinite loop spaces $X$ and $Y$, $X \otimes Y := \Omega^\infty (X_0 \wedge Y_0)$, for $X_0$ and $Y_0$ such that $\Omega^{\infty} X_0 \cong X$ and $\Omega^{\infty} Y_0 \cong Y$.

For instance, in the case for $QX \otimes QY$, this would be the same as $\Omega^{\infty} (Q(S^{\infty} \wedge X) \wedge Q(S^{\infty} \wedge Y))$.

Also, it would seem that $\Omega^{\infty}$ and $\Sigma^{\infty} := S^{\infty} \wedge -$ form an adjunction between spaces and infinite loop spaces.

My question is:

  1. Is it true that QX is homotopic to $(\Omega^{\infty} \circ \Sigma^{\infty})X$ for X a based CW complex?
  2. If not, ss it true that $(\Omega^{\infty} \circ \Sigma^{\infty})X$ is an infinite loop space?
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  • $\begingroup$ As far as I can see, there is no question being asked here. Also, what exactly is the space S^∞? It would be best to give references for the source you are using. $\endgroup$
    – Dmitri Pavlov
    Commented Feb 18 at 23:08
  • $\begingroup$ Or at least define your notation. (Is S^∞ the increasing union of the n-spheres S^n ?) $\endgroup$
    – Daniel Asimov
    Commented Feb 19 at 0:23
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    $\begingroup$ Do you know that $S^\infty$ is contractible? So infinite loop spaces in your sense are always contractible. $\endgroup$
    – Dan Petersen
    Commented Feb 19 at 6:17
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    $\begingroup$ $\Sigma^\infty$ is not usually defined as $S^\infty\wedge -$. Rather, $\Sigma^\infty$ is a functor from spaces to spectra, which sends $X$ to $\{S^n X\mid n=0,1, \ldots\}$. $\Sigma^\infty, \Omega^\infty$ define an adjunction of spaces with spectra, rather than with infinite loop spaces. You do have an adjunction between spaces and infinite loop spaces: ($\Omega^\infty \Sigma^\infty$, inclusion). But $\Omega^\infty \Sigma^\infty$ is not Map$(S^\infty, S^\infty \wedge -)$. Rather, it is (ho)colim$_n \Omega^n S^n -$. The distinction is very important. $\endgroup$
    – Gregory Arone
    Commented Feb 19 at 8:53
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    $\begingroup$ @GregoryArone sounds like you can show that 1 is false. $\endgroup$
    – user30211
    Commented Feb 19 at 13:49

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