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I apologize in advance if this question is too technical. I haven't found a reference in the literature yet, and it seems difficult enough that perhaps it has not been answered.

Let $A$, $B$, and $C$ be based topological spaces, all compactly generated and weak Hausdorff. Let $\wedge$ and $F$ denote smash product and space of based functions, respectively, modified in the usual way so as to produce CGWH spaces. Consider the "assembly" map $$ A \wedge F(B,C) \rightarrow F(B, A \wedge C) $$ It is the adjoint of the smash product of the identity of $A$ and the evaluation map of $F(B,C)$.

Is this map always a closed inclusion?

(It is not hard to show that it is injective, has closed image, and that if $K$ is a closed subset of the domain, then $K \cup \{*\}$ is a closed subset of the codomain. If you take $A$, $B$, and $C$ to be $[0,\infty)$, you appear to get a counterexample.)

EDIT: I believed earlier that this counterexample could maybe be fixed by making everything compactly generated, but that doesn't help, you can still produce a non-convergent sequence in the source that converges in the target to the basepoint, and this still produces a closed set whose image is not compactly closed. So this is actually not true! Juan, thanks for pointing me to Lewis's thesis.

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Isn't this addressed in Lewis' Appendix A (on compactly generated spaces)? His Proposition 8.5 gives a criteria (for A compact), to get a closed inclusion. But your comment on adding the disjoint basepoint is important ("The adjunction od the disjoing basepoint is essential to the [...] result"). See his CounterExample 8.6 following the proposition.

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  • $\begingroup$ Hi Juan, you're right, he does address it, and specifically shows that the answer to my question is no. I'm interested in the case where $A$ is the $n$-sphere, and it looks like I should expect this map to not be a closed inclusion. Too bad... I wanted to say that $\Sigma^\infty_+ LX$ is homeomorphic to the totalization $Tot(\Sigma^\infty_+ X^{\bullet + 1})$, but it looks like that's not true now. $\endgroup$
    – Cary
    Commented Apr 14, 2015 at 13:19

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