Every finite-dimensional Hopf algebra is isomorphic to a dual Hopf algebra? 1. Context
Let $H$ be a Hopf algebra over a field $\mathbb{k}$. Denote by $I_l(H)$, $I_r(H)$ its space of left integrals/right integrals respectively.
I am studying a proof of the following proposition:

If $H$ is finite dimensional, then $\dim(I_l(H)) = \dim(I_r(H)) = 1$.

The proof seems to use the following lemma:

For any finite dimensional Hopf algebra $H$ there exists a Hopf algebra $M$ such that $M^*$ and $H$ are isomorphic as Hopf algebras.

2. Question
Why does the lemma hold?
3. A bit more context, if needed
The proof continues as follows:

Due to that lemma it suffices to show that for any finite-dimensional Hopf algebra $M$ its dual Hopf algebra $M^*$ satisfies $\dim(I_l(M^*)) = \dim(I_r(M^*))=1$.
One endows the vector space $M^*$ with the structure of a (right) $H$-Hopf-module (with the lower harpoon action, and a certain coaction $\Delta_{M^*}$). Using the Fundamental theorem of Hopf modules as well as the fact that $M^*$ is finite dimensional, one obtains $\dim((M^*)^{\mathrm{coH}})=1$. Here, $(M^*)^{\mathrm{coH}} = \{x \in M^* : \Delta_{M^*}(x) = x \otimes 1\}$ denotes the space of right coinvariants. Using finite-dimensionality again, a calculation shows that $(M^*)^{\mathrm{coH}} = I_l(M^*)$. Analogously, one proves that $\dim(I_r(M^*)) = 1$.

 A: We have for every vector space $V$ a natural homomorphism
$$
  φ_V
  \colon
  V \longrightarrow V^{**} \,
  \quad
  v \longmapsto [v^* \mapsto v^*(v)] \,.
$$
If $V$ is finite-dimensional, then $φ_V$ is an isomorphism.
If $H$ is a finite-dimensional Hopf algebra, then $φ_H$ is not just an isomorphism of vector spaces, but an isomorphism of Hopf algebras.
Therefore,
$$
  H ≅ H^{**}
$$
as Hopf algebras.
Let us check the above claim that $φ_H$ is an isomorphism of Hopf algebras.
It remains to show that $φ_H$ is a homomorphism of Hopf-algebras.
We recall how the dual of a Hopf algebra is constructed:

Let $H$ be a finite-dimensional Hopf algebra.
The multiplication of $H^*$ is defined by
$$
  (h^*_1 h^*_2)(h) = \sum_{(h)} h^*_1( h_{(1)} ) \, h^*_2( h_{(2)} ) \,.
$$
The unit of $H^*$ is precisely the counit of $H$ (which is a linear map from $H$ to $$, and therefore an element of $H^*$), i.e., $1_{H^*} = ε_H$.
The comultiplication of $H^*$ is implicitly determined by
$$
  h^*(h_1 h_2)
  =
  \sum_{(h^*)} h^*_{(1)}(h_1) \, h^*_{(2)}(h_2) \,.
$$
The counit of $H^*$ is given by evalution at the unit of $H$, i.e., $ε_{H^*}(h^*) = h^*(1_H)$.
The antipode of $H^*$ is given by the dual map of the antipode of $H$, i.e., $S_{H^*} = S_H^*$.

Let us now show that $φ$ is a homomorphism of bialgebras:

*

*The chain of equalities
\begin{align*}
  φ(h_1 h_2)(h^*)
  =
  h^*(h_1 h_2)
  =
  \sum_{(h^*)} h^*_{(1)}(h_1) \, h^*_{(2)}(h_2)
  &=
  \sum_{(h^*)} φ(h_1)(h^*_{(1)}) \, φ(h_2)(h^*_{(2)}) \\
  &=
  (φ(h_1) φ(h_2))(h^*)
\end{align*}
shows that $φ$ is multiplicative.


*That $φ$ preserves units follows from $φ(1_H)(h^*) = h^*(1_H) = ε_{H^*}(h^*) = 1_{H^{**}}(h^*)$.


*The chain of equalities
$$
  φ(h)(h^*_1 h^*_2)
  =
  (h^*_1 h^*_2)(h)
  =
  \sum_{(h)} h^*_1( h_{(1)} ) \, h^*_2( h_{(2)} )
  =
  \sum_{(h)} φ(h_{(1)})(h^*_1) \, φ(h_{(2)})(h^*_2)
$$
shows that the element $\sum_{(h)} φ(h_{(1)}) ⊗ φ(h_{(2)})$ satisfies the defining equality of the element $Δ_{H^{**}}(φ(h))$.
In other words, $φ$ is comultiplicative.


*The equalities $ε_{H^{**}}(φ(h)) = φ(h)(1_{H^*}) = 1_{H^*}(h) = ε_H(h)$ show that $φ$ preserves counits.
This shows that $φ$ is a homomorphism of bialgebras, and therefore (automatically) a homomorphism of Hopf algebras.
