I am reading a little about cobordism and I have a basic question, which makes sense both in the topological and motivic setting. Let $\mathrm{Gr}\_{n,\infty}$ denote the infinite $n$-Grassmanian and denote $\xi\_n $ its universal bundle. The cobordism spectrum $\mathrm{MGL}$ is made out of the universal Thom spaces.That is to say, it is the spectrum
$$
(S\, ,\, \mathrm{Th}(\xi\_1)\, ,\, \mathrm{Th}(\xi\_2)\, ,\, \ldots \, ,\, \mathrm{Th}(\xi\_i)\, ,\, \ldots \ )
$$
with the natural bounding maps. My question is



> 
> Is $\mathrm{MGL}[2r](r)$ isomorphic, in $\mathbf{SH}$, to the spectrum $(\, \mathrm{Th}(\xi\_r)\, ,\, \mathrm{Th}(\xi\_{r+1})\, ,\, \ldots \, ,\, \mathrm{Th}(\xi\_{r+i})\, ,\, \ldots \ )$? And if so, why?
> 
> 
> 


It seems to me that this should be right and simple, but since $\mathrm{MGL}$ is not an $\Omega$-spectrum I fail to see a direct reason. 



Basic question on the cobordism spectrum
A simple way of seeing it is to explicitly spell out what we mean when we say that a spectrum is "presented" by a prespectrum. To say that that a spectrum $E$ is presented by
$$(E\_0,E\_1,...)$$
means that, in whatever model for motivic spectra you are using,
$$E\cong \mathrm{colim}\_k \Sigma^{-2k,-k}\Sigma^{\infty}E\_k$$
(here the colimit is a homotopy colimit). Since suspensions commute with colimits (being equivalences) we have that
$$\Sigma^{2n,n}E\cong \mathrm{colim}\_k \Sigma^{2(n-k),(n-k)}\Sigma^{\infty}E\_k\cong \mathrm{colim}\_k \Sigma^{-2k,-k}\Sigma^\infty E\_{k+n}$$
where the last step is just reindexing the colimit. In particular, $\Sigma^{2n,n}E$ is presented by
$$(E\_n,E\_{n+1},\dots)\,.$$


