(Algebraic) cobordism and the rank function

Question

I write the question for algebraic cobordism but I have the analogue question for classic cobordism.

The spectrum representing algebraic cobordism $$ \mathbf{MGL}=(*, \mathrm{Th}(1) , \ldots , \mathrm{Th}(n), \ldots) $$ is given by the Thom spaces $\mathrm{Th}(n)$ of the universal bundles of the Grassmanians $\mathrm{Gr}_n$. Let $\mathbf{E}=(E_n)_{n\in\mathbb{N}}$ be the spectrum of a cohomology theory with Chern classes (or oriented), it is very easy to write the universal map $\varphi^E\colon \mathbf{MGL}\to \mathbf{E}$ since it is the morphism of spectra having at the $n$-th level the Thom class $\mathit{th}^\mathbf{E}(\mathrm{Th}(n)) :\mathrm{Th}(n)\to E_n$.

Now, for the concrete case of motivic cohomology (or singular cohomology in the topological case) one can check that at the (0,0)-level the map $$ \mathbf{MGL}^{0,0}(X)\to H^{0}(X,\mathbb{Z}(0)) $$ induced by $\varphi^\mathbf{H}$ "is" the rank function. More concretely, the above arrow is the composition of the following: $$ \mathbf{MGL}^{0,0}(X)\xrightarrow{\varphi^K} K_{0}(X)\xrightarrow{\mathrm{rank}}H^{0}(X,\mathbb{Z}(0)) $$ One can check this explicitly, for example, whenever you know that $\mathbf{MGL}^{0,0}(X)$ is generated by morphisms $f\colon Y\to X$ where $\mathrm{dim} Y=\mathrm{dim}X$. This suggests that the unstable map $$ (\varphi^H)_0: \Omega^\infty \mathbf{MGL}\to K(0,0) $$ defined as the adjoint of the composition in $\mathbf{SH}$ of $\Sigma^\infty \Omega^\infty \mathbf{MGL}\to \mathbf{MGL}\xrightarrow {\varphi^H} \mathbf{H}$, should equal the composition in the unstable homotopy category of $$ \Omega^\infty \mathbf{MGL} \xrightarrow{(\varphi^K)_0}\mathbb{Z}\times \mathrm{Gr}\xrightarrow{\mathrm{rank}} K(0,0), $$ where $(\varphi^K)_0$ is the defined analogously to $(\varphi^H)_0$ and $K(0,0)$ is the space at level zero of the spectrum $\mathbf{H}$ representing motivic cohomology. To sum up, my question is:

Does $(\varphi^H)_0$ equal to $\mathrm{rank}\circ (\varphi^K)_0$?

On top of the motivation I gave above and to rephrase the question. It is already known that, over a field, $\Omega^\infty \mathbf{MGL}=\mathbb{Z}\times\mathrm{Hilb}_\infty^{\mathrm{lci}}(\mathbb{A}^\infty)^+$ (see [1]) and the rank function on $K$-theory is given by the projection towards the first factor of $\mathbb{Z}\times \mathrm{Gr}$, so it is very likely that $(\varphi^H)_0$ is also the projection towards the first factor composed with its natural map towards $K(0,0)$.

If you know any reference of a computation similar to this for classic cobordism I would also thank that.

Answer 1

Let me first write what happens for classical cobordism. You are basically asking whether the map $\operatorname{MU}\to H\mathbb{Z}$ factors through the projection $\operatorname{ku}\to H\mathbb{Z}$. But this is clear, since all spectra in sight are connective and we have an equivalence $$\operatorname{Map}(E,H\mathbb{Z})\cong \operatorname{Map}(\pi_0E,\mathbb{Z})$$ for every connective spectrum $E$ given by the fact that $\pi_0$ is the left adjoint of the inclusion of discrete spectra (i.e. abelian groups) into connective spectra. This adjunction by the way works both in spectra and abelian groups and $E_\infty$-ring spectra and commutative rings, in which case it's clear that the right hand side is just a point both for $\operatorname{MU}$ and $\operatorname{ku}$.

Something similar happens in motivic cohomology. Here the statement is that the map $\operatorname{MGL}\to H\mathbb{Z}$ factors uniquely through the orientation $\operatorname{MGL}\to\operatorname{kgl}$ (here with $\operatorname{kgl}$ we mean the very effective cover of the motivic spectrum $\operatorname{KGL}$). Instead of taking homotopy groups the correct thing to do is taking slices. Indeed the both map $\operatorname{MGL}\to H\mathbb{Z}$ and $\operatorname{kgl}\to H\mathbb{Z}$ exhibit $H\mathbb{Z}$ as the zero slice of the respective spectrum and $\operatorname{MGL}\to\operatorname{kgl}$ induces an equivalence on 0-slices.