TITLE: Is there a map of spectra implementing the Thom isomorphism?
QUESTION [23 upvotes]: A well known theorem in algebraic topology relates the (co)homology of the Thom space $X^\mu$ of a orientable vector bundle $\mu$ of dimension $n$ over a space $X$ to the (co)homology of $X$ itself: $H_\ast(X^\mu) \cong H_{\ast-n}(X)$ and $H^\ast(X^\mu) \cong H^{\ast-n}(X)$.
This isomorphism can be proven in many ways: Bott & Tu has an inductive proof using good covers for manifolds and I learned on MathOverflow that one can use a relative Serre spectral sequence. However, I believe that there should also be a proof using stable homotopy theory, in the case of homology by directly constructing a isomorphism of spectra $X^\mu \wedge H\mathbb{Z} \to X_+ \wedge \Sigma^{-n} H\mathbb{Z}$, where $X^\mu$ denotes the Thom spectrum, $H\mathbb{Z}$ the Eilenberg-Mac Lane spectrum for $\mathbb{Z}$ and $X_+$ the suspension spectrum of $X$ with a disjoint basepoint added.
Is there an explicit construction of such a map implementing the Thom isomorphism on the level of spectra? I am interested in such a construction for both homology and cohomology. If so, is there a similar construction for generalized (co)homology theories? I would also be interested in references.

REPLY [3 votes]: I just noticed this question, so my apologies for a very belated answer.  Proposition 20.5.5 of http://www.math.uchicago.edu/~may/EXTHEORY/MaySig.pdf states that a $k$-orientation of a spherical fibration $X$ over a base space $B$ specifies a  $k$-trivialization of $X$.  Here $k$ is any (homotopy) commutative ring spectrum, and the notion of a $k$-orientation is the standard one.  However, a $k$-trivialization 
is an equivalence  $k\wedge X\simeq k\wedge S^n_B$ of parametrized spectra over $B$. Intuitively, after smashing with $k$ over $B$, $X$ thinks that it is the trivial fibration $B \times S^n \to B$.The Thom isomorphism follows directly.