# How to prove that $MU$ is an oriented spectrum? A doubt in the proof in Kochman's book

I want to show that the Thom spectrum $MU$ is oriented, namely I want to find a class $x \in \widetilde{MU}^2(\mathbb{C}P^{\infty})$ whose restriction to $\widetilde{MU}^2(S^2)$ is a generator. in his book, Kochman claims that (page 132) $$\xi \colon \mathbb{C}P^{\infty} \simeq MU(1) \xrightarrow{Id} MU(1)$$ is the orientation.

I don't get why this map is an orientation, since (by def.) it should be an element in $\widetilde{MU}^2(\mathbb{C}P^{\infty})$ i.e. (up to homotpy) a map $$\Sigma^{\infty}\mathbb{C}P^{\infty} \to \Sigma^2 MU$$ which is uniquely determined by a map $$\mathbb{C}P^{\infty} \to MU(2)$$

I don't get how to interpret the proof of the author since it's unclear to me why the identity comes into play.

I think you just forgot that $\mathbb{C}$ is two dimensional. Is $MU(n)$ in your notation the Thom space of the universal rank $n$ complex vector bundle on $BU(n)$?
If so, because $\mathbb{C}$ has dimension two, the structure maps of the spectrum $MU$ go $\Sigma^2 MU(n) \to MU(n+1)$, which means that $MU(1) \mapsto \Omega^{\infty-2} MU$, not $\Omega^{\infty-1} MU$. So a map $X \to MU(1)$ gives a class in $\widetilde{MU}^2(X)$.