# Basic question on the cobordism spectrum

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.

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 $$(E_n,E_{n+1},\dots)\,.$$