Finding the expression for $f_h^{-1} (a)$ when the expression for $f_h (a)$ is given. I am following the section $6.1$ on Deformations of Hopf Algebras (Chapter $6$) from A Guide to Quantum Groups written by Chari and Pressley. Let $A$ be a Hopf algebra over $k$ with two deformations $A_h$ and $A_h'$. Then they are said to be equivalent if there exists a $k[[h]]$ module isomorphism $f_h : A_h \longrightarrow A_h'$ such that $f_h \equiv \text {id}\ (\text {mod}\ h).$ The last modularity relation means that $f_h$ acts as identity on the elements of $A.$ So there exist $k[[h]]$ module homomorphisms $f_i : A_h \longrightarrow A_h'$ such that $$f_h(a) = a + h f_1 (a) + h^2 f_2 (a) + \cdots$$
From here the authors made a note (Page no. $172$) on the following equality $:$ $$f_h^{-1} (a) = a - h f_1 (a) + \cdots$$
But I can't get it. Could anyone please shed some light on it?
Thanks for your time.
 A: The $f_i$ are not maps $A_h \to A'_h$, but instead maps $A \to A$.
These maps arise as follows:

*

*We may assume for simplicity that $A_h, A'_h = A⟦h⟧$, since we only care about the $⟦h⟧$-module structures of $A_h$ and $A'_h$ for the moment.


*The equivalence $f_h \colon A_h \to A'_h$ is in particular $⟦h⟧$-linear, and therefore a $⟦h⟧$-linear map from $A⟦h⟧$ to $A⟦h⟧$.


*We regard $A$ as the constant power series in $A⟦h⟧$.
The map $f_h$ is then uniquely determined by its restriction $f_h|_A$; more explicitly
$$
  f_h\Biggl( \sum_{i = 0}^∞ a_i h^i \Biggr)
  =
  \sum_{i = 0}^∞ (f_h|_A)(a_i) h^i \,.
$$
So how does this restriction $f_h|_A$ look like? Well, it is a $$-linear map from $A$ to $A⟦h⟧$.
It is therefore of the form
$$
  (f_h|_A)(a) = \sum_{i = 0}^∞ f_i(a) h^i
$$
for unique maps $f_0, f_1, \dotsc$ from $A$ to $A$.
It follows from the $$-linearity of $f_h|_A$ that the maps $f_i$ are again $$-linear.


*Very explicitly, we have
$$
  f_h\Biggl( \sum_{i = 0}^∞ a_i h^i \Biggr)
  = \sum_{i = 0}^∞ f_h(a_i) h^i
  = \sum_{i = 0}^∞ \sum_{j = 0}^∞ f_j(a_i) h^j h^i
  = \sum_{i, j = 0}^∞ f_i(a_j) h^{i + j} \,.
$$


*The condition $f_h ≡ \mathrm{id}_A$ modulo $h$ tells us that $f_0 = \mathrm{id}_A$.
We can similarly consider the restriction $f_h^{-1}|_A$, which is of the form
$$
  (f_h^{-1}|_A)(a) = \sum_{i = 0}^∞ f'_i(a) h^i
$$
for unique $$-linear maps $f'_0, f'_1, \dotsc,$ from $A$ to $A$, and with $f'_0 = \mathrm{id}_A$.
The $⟦h⟧$-linear map $f_h^{-1}$ is given in terms of its restriction $f_h^{-1}|_A$ by
$$
  f_h^{-1}\Biggl( \sum_{i = 0}^∞ a_i h^i \Biggr)
  =
  \sum_{i, j = 0}^∞ f'_i(a_j) h^{i + j} \,.
$$
The functions $f'_i$ are determined by $f'_h$, which is determined by $f_h$, which in turn is determined by the maps $f_i$.
We should therefore be able to express the functions $f'_i$ in terms of the functions $f_i$.
We observe that
$$
  \sum_{i = 0}^∞ a_i h^i
  =
  f'_h\Biggl( f_h\Biggl( \sum_{i = 0}^∞ a_i h^i \Biggr) \Biggr)
  =
  \sum_{i, j, k = 0}^∞ f'_i( f_j( a_k ) ) h^{i + j + k} \,.
$$
We can now compare coefficients:

*

*In degree $0$ we already know that $f'_0, f_0 = \mathrm{id}_A$.

*In degree $1$ we have
$$
  a_1 h
  = f'_1(f_0(a_0)) h + f'_0(f_1(a_0)) h + f'_0(f_0(a_1)) h
  = f'_1(a_0) h + f_1(a_0) h + a_1 h \,,
$$
and therefore $f'_1(a_0) = -f_1(a_0)$.
This shows that $f'_1 = -f_0$.

*In degree $2$ we have
\begin{align*}
  a_2 h^2
  &= f'_2(a_0) h^2 + f_2(a_0) h^2 + a_2 h^2 + f'_1(f_1(a_0)) + f'_1(a_1) h^2 + f_1(a_1) h^2
  \\
  &= [ f'_2(a_0) + f_2(a_0) - f_1(f_1(a_0)) - f_1(a_1) + f_1(a_1) + a_2 ] h^2
  \\
  &= [ f'_2(a_0) + f_2(a_0) - f_1(f_1(a_0)) + a_2 ] h^2
\end{align*}
It follows that
$$
  f'_2(a_0) = f_1(f_1(a_0)) - f_2(a_0) \,.
$$
We can continue in this way to express the maps $f'_i$ in terms of the maps $f_i$.
