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Let $X$ be an $n$-connective spectrum for some $n\in \mathbb{Z}$. Is then $[X, Y] = [X, Y\langle n\rangle]$ for all spectra $Y$, where $Y\langle n\rangle$ denotes the $n$-connective cover of $Y$?

Dually, if $Y$ is $n$-truncated, i.e. $\pi_k(Y) = 0$ for all $k > n$, is then $[X, Y] = [X_{\leq n}, Y]$ for all spectra $X$, where $X_{\leq n}$ denotes the truncation of $Y$?

In other words, are taking the connected cover respectively truncation a right respectively left adjoint?

What about the mapping spectra?

\edit: I mixed up the second part as noted in the comments.

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    $\begingroup$ Yes, no (you've got it backwards), yes, and no (because mapping spectra know about maps from desuspensions of the source.) EDIT: Now you've got the second thing the right way round, so yes to that too. $\endgroup$
    – Dylan Wilson
    Commented Sep 25, 2017 at 21:04
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    $\begingroup$ The mapping spectrum will usually not be n-connective period. After all, the mapping spectrum into the sphere flips your spectrum upside down... or more readily: mapping spectra out of a sphere will give you desuspensions of the target, which are evidently less connective than they started. $\endgroup$
    – Dylan Wilson
    Commented Sep 25, 2017 at 21:10
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    $\begingroup$ Yeah all your statements are okay with mapping space replacing [,]. (Obligatory comment about using derived mapping spaces if you're in some model category.) $\endgroup$
    – Dylan Wilson
    Commented Sep 25, 2017 at 21:11
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    $\begingroup$ Just to tag along with Dylan's comments, if you want bounds on the nonzero groups of the function spectrum you most easily get them from bounds on the homotopy groups of Y and the cohomology groups of X. Connectivity can give you a bound on the cohomology of X but coconnectivity can't. $\endgroup$
    – Tyler Lawson
    Commented Sep 26, 2017 at 1:00
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    $\begingroup$ Of course, then there are fun examples like $X = \mathbb RP^{\infty}$ and $Y = S$ (Segal conjecture for $\mathbb Z/2$) to mess with your head! (The mapping spectrum is still $-1$-connected.) $\endgroup$
    – Nicholas Kuhn
    Commented Sep 26, 2017 at 3:06

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