In a comment for this old [question](https://mathoverflow.net/questions/92081/hopf-comodule-decompositions-and-coinvariant-elements), it was said that 


>
There is a theorem that Hopf modules are (up to isomorphism) the tensor products of their coinvariants with H. (This theorem is usually worded in terms of a category equivalence. I don't know a good reference.) 


I'm guessing that this means that given a (right) $H$-comodule $V$, for $H$ a Hopf algebra, equipped with a right $H$ action for which $\Delta\_R(vh) = v\_{(0)} g\_{(0)} \otimes v\_{(1)}g\_{(1)}$, we have an isomorphism
$$
V \simeq V\_{\text{inv}} \otimes H
$$
Now it's easy to see that we have a surjective map
$$
 V\_{\text{inv}} \otimes H \to V, ~~~ v \otimes h \mapsto vh.
$$
How does one show that this is an isomorphism?


EDIT: Wait, I think this is actually obvious: The map
$$
V \to V\_{\text{inv}} \otimes H, ~~~~~ v \mapsto v\_{(0)} S(v\_{(1)}) \otimes v\_{(2)},
$$
seems to have the multiplication map as its inverse. So this gives us the isomorphism. Yes?



$H$-Hopf modules equal the tensor products of their coinvariants with H
The result you mention is a classical result on Hopf modules, first proved by Larson and Sweedler. My favorite reference is 


Pareigis: When Hopf Algebras are Frobenius Algebras. J. of Alg. 18(1971), 588-596. Lemma 2. 


There you can also find the maps you describe in your question (so the answer is yes, they are inverse to each other). 


Another reference is Sweedler's book (mentioned already by Mariano), Theorem 4.1.1. However, I think in the book the base ring is always a field, while Pareigis works over a comm. ring. 


Somewhere in the book "Brzezinski, Wisbauer: Corings and Comodules" I read that the result in question also follows from a more general theorem on comodules over corings. But I don't know details. 


