Let $X$ be a fixed parametrizing space. Let $E$ and $E'$ be two spectra and let $E_X$ and $E'_X$ be their trivial parametrized versions. Intuitively I imagine that the morphisms of parametrized spectra $E_X\to E'_X$ should correspond to maps $X\to \operatorname{Map}(E, E')$ where the mapping space on the right should be the underlining infinite loop space of the function spectrum, if you wish $\operatorname{Map}(E, E')=\Omega^\infty F(E, E').$

Is this, or at least anything similar to it, actually true?


This is true.

To prove it I will use the fact that maps of parametrized spectra can be computed as natural transformations of functors from $X$ into the $\infty$-category of spectra (cfr. this paper) and the formula for computing the space of natural transformation (e.g. see here, proposition 2.3):

$$\mathrm{Map}(E_X,E'_X) = \int_{x\in X} \mathrm{Map}(E_x,E'_x) = \int_{x\in X} \mathrm{Map}(E,E') = \mathrm{Map}(\tilde O(X), \mathrm{Map}(E,E')) = \mathrm{Map}(X,\mathrm{Map}(E,E'))$$

Here $\tilde O(X)$ is the twisted arrow category of $X$, seen as an $\infty$-groupoid, which is just equivalent to $X$ (one way to see this is that the canonical projection $\tilde O(X)\to X$ is a left fibration with contractible fibers).

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    $\begingroup$ In fact you can say more: if $E$ is any parametrized spectrum, $\mathrm{Map}(E,E'_X) = \mathrm{Map}(\mathrm{colim}_x E_x, E')$ (with exactly the same proof as above). $\endgroup$ – Denis Nardin Oct 3 '16 at 13:38
  • $\begingroup$ Hey, I asked a follow-up question about the possibility of extending this to (commutative) ring spectra. If you have the time, I would be very grateful if you would take a look at that as well. Cheers! $\endgroup$ – A Rock and a Hard Place Oct 4 '16 at 19:45

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