Are there non trivial maps from $H\mathbb{Z}$ to $MGL$? Let $k$ be a field of characteristic $0$. Let us denote by $\mathbf{1}_{k}$ the sphere spectrum. Let $MGL$ be the algebraic cobordism spectrum. 
We have the following diagram 
$$H\mathbb{Z}\leftarrow \mathbf{1}_{k}\rightarrow MGL$$
My question is the following:

Are there non trivial maps $H\mathbb{Z}\to MGL$ such that the above triangle commutes?

 A: NO such a map does not exist.
Thanks to Eric Peterson for making me realize that the argument carries through even if the map is not a map of algebras.
By rigidity,you can only consider the case where $k$ is a subfield of $\mathbb{C}$, so I will restrict myself to this case.

If there were such a map, then it would exist also after Betti realization. Hence there would be a map of spectra $H\mathbb{Z}→MU$ such that the precomposition with the unit $\mathbb{S}→H\mathbb{Z}$ is the unit $\mathbb{S}→MU$. But then, postcomposing with the standard cotruncation map $MU→H\mathbb{Z}$ (this is just an avatar of the ring map $E→Hπ_0E$ existing for every connective ring spectrum) we would have a map $H\mathbb{Z}→H\mathbb{Z}$ that is the unit when precomposed with $\mathbb{S}→H\mathbb{Z}$. Since endomorphisms of $H\mathbb{Z}$ are determined by the value on $\pi_0$, it follows that the composition $H\mathbb{Z}→MU→H\mathbb{Z}$ must be the identity. In particular $H\mathbb{Z}$ would be a retract of $MU$.
However this is false: in fact $MU→H\mathbb{Z}$ factors through connective $K$-theory $ku$, so in particular the bottom cell of $ku$ would split off. But we know that the first k-invariant of $ku$ is nontrivial (it is in fact the Milnor invariant).
