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To fix the notation, a spectrum is a sequence of spaces indexed over $\mathbb{N}$ (!) with structure maps as usual.

My first questions is: Does the suspension spectrum functor embed the category of spaces into spectra? It is clearly a faithful functor and injective on objects but I doubt that it is full.

There is a unstable model structure on spectra by defining weak equivalences and fibrations degreewise.

My second question is: Does the suspension spectrum functor embed the homotopy category of spaces into the unstable homotopy category of spectra?

Define two endofunctors $s$ and $t$ on the category of spectra by setting $sX_n=X_{n+1}$ and $tX_n=X_{n-1}$ where $X_{-1}=*$. Then $t$ is left adjoint to $s$ and this is a Quillen adjunction with respect to the unstable model structure.

My third question is: Is it true that $s$ is left adjoint to $t$, too? If yes, is this also a Quillen adjunction? I have checked it and it seems to be right but somehow I feel not well about it.

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The answer to (1) is that the suspension-spectrum functor from based spaces to spectra is fully faithful. Given spaces X and Y, a map between their suspension spectra $\Sigma^\infty X \to \Sigma^\infty Y$ is a collection of maps $S^n \wedge X \to S^n \wedge Y$ respecting the structure maps. In particular, taking $n=0$ we recover the map $X \to Y$, and since the structure maps are all isomorphisms any map on level zero extends uniquely to the rest of the spectrum. Another way to see this is that the suspension spectrum functor is left adjoint to the functor $U$ that takes a spectrum $\{E_n\}$ to $E_0$.

The answer to (2) is that the suspension spectrum functor is faithful; it has a left inverse, which is obtained from pushing the functor $U$ down to the homotopy categories.

The answer to (3) is that s is not left adjoint to t on the level of spectra For instance, one has $s(\Sigma^\infty X) \cong \Sigma^\infty(S^1 \wedge X)$ for any $X$, and so by the adjunction mentioned in (1), we have: $$ Hom(s \Sigma^\infty X, \Sigma^\infty Y) = Hom(S^1 \wedge X, (\Sigma^\infty Y)_0) = Hom(X,Y) $$ However, we also have: $$ Hom(\Sigma^\infty X, t \Sigma^\infty Y) = Hom(X,(t\Sigma^\infty Y)_0) = * $$

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I'm a bit confused: I thought one of the points of the stable category is that you could define maps on cofinal subspectra... so that, in particular, there may not be a map defined at n=0. Or am I confusing "maps" and "functions" or something? –  Dylan Wilson Aug 10 '11 at 6:08
    
@Dylan: I believe you're talking about the definition as presented by Adams ("cells now, maps later"). One of the chief reasons to use that definitions is so that a homotopy class of map $E \to F$ can always be lifted to one of these "cofinal" sequences of maps. However, my understanding is that this proved to be too difficult to work with in general. Almost all modern categories of spectra (such as the version in Bousfield and Friedlander) define maps to be "on the nose" and use weak equivalences/model categories to define the homotopy category. –  Tyler Lawson Aug 10 '11 at 19:00
    
...I assumed that the questioner was using one of these definitions because they were talking about "unstable" spectra and a model structure on those. –  Tyler Lawson Aug 10 '11 at 19:00
    
Dear Tyler, thanks for the effort. Doesn't the fact that $U\circ \Sigma^\infty=id$ prove that $\Sigma^\infty$ is injective on objects and morphism sets only? I have problems seeing that a morphism $f:X\to Y$ extends uniquely to $\Sigma f:\Sigma X\to\Sigma Y$. Why? I can follow your argument for (3) (maybe you mean $hom(S^1\wedge X,Y)$ in the first row) but where is my mistake in this observation: Why can't I extend a map $X\to ∗$ in degree $0$ and $\Sigma X\to Y$ in degree $1$ to a map $\Sigma^\infty X→t\Sigma^\infty Y$? The condition of a spectrum map in deg $0$ is then fullfilled! –  John Zhao Aug 11 '11 at 8:00
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@John: If one writes $\sigma_n:\Sigma X_n \to X_{n+1}$ for the structure maps, the definition of a map of spectra is a collection $f_n:X_n \to Y_n$ such that $f_{n+1} \circ \sigma_n = \sigma_n \circ (\Sigma f_n)$. This is more than asking $f_{n+1}$ to be an extension of $f_n$, and if $\sigma_n$ are all isomorphisms on the domain one is forced into $f_{n+1} = \sigma_n (\Sigma f_n) \sigma_n^{-1}$. –  Tyler Lawson Aug 11 '11 at 12:20

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