Finite Hopf algebras have one-dimensional integral spaces and bijective antipode

Apply [F2] and [F3] to $K=H$. The action map ${\int }_{H^{\ast }}^{\mathrm{\ell }}\otimes H\to H^{\ast }$ is an isomorphism, so $(\dim {\int }_{H^{\ast }}^{\mathrm{\ell }})(\dim H)=\dim H^{\ast }$. Since $0<\dim H=\dim H^{\ast }$ by [F8], division by this positive integer gives $\dim {\int }_{H^{\ast }}^{\mathrm{\ell }}=1$; in particular this integral space contains a nonzero $\lambda$.

Apply [F2]–[F3] to $K=H^{\ast }$, which is a finite-dimensional Hopf algebra by [F4]. Its Hopf-module isomorphism gives $(\dim {\int }_{H^{\ast \ast }}^{\mathrm{\ell }})(\dim H^{\ast })=\dim H^{\ast \ast }$; applying [F8] to $H$ and $H^{\ast }$ gives $\dim H^{\ast \ast }=\dim H^{\ast }$, so $\dim {\int }_{H^{\ast \ast }}^{\mathrm{\ell }}=1$. Choose a nonzero $\mu \in {\int }_{H^{\ast \ast }}^{\mathrm{\ell }}$. If $S_{H^{\ast }}(f)=0$, the action formula [F2] gives $\mu \leftharpoonup f=0$; injectivity of the Hopf-module isomorphism forces $\mu \otimes f=0$. Since $\mu$ is a basis of the one-dimensional space, its dual basis functional takes $\mu$ to $1$ by [F8], and applying that functional to the first tensor factor gives $f=0$. Thus $S_{H^{\ast }}$ is injective and hence bijective by [F7]. The evaluation algebra and coalgebra isomorphism [F5] carries the left-integral equations between $H$ and $H^{\ast \ast }$, so $\dim {\int }_H^{\mathrm{\ell }}=1$.

If $S_H(h)=0$, then $(\lambda \leftharpoonup h)(x)=\lambda (x{S}_H(h))=0$ for every $x$, so the isomorphism in step 1.1 sends $\lambda \otimes h$ to zero. Its injectivity gives $\lambda \otimes h=0$. Since $\lambda$ is a basis of the one-dimensional space ${\int }_{H^{\ast }}^{\mathrm{\ell }}$, its dual basis functional $q$ satisfies $q(\lambda )=1$ by [F8]; applying $q\otimes {\mathrm{i}\mathrm{d}}_H$ yields $h=0$. Thus $S_H$ is injective, and [F7] makes it bijective.

For either $K=H$ or $K=H^{\ast }$, bijectivity of $S_K$ and anti-multiplicativity show that $S_K$ sends left integrals to right integrals: for $t\in {\int }_K^{\mathrm{\ell }}$ and $a\in K$, apply $S_K$ to $S_K^{-1}(a)t={\epsilon }_K(a)t$ and use ${\epsilon }_K{S}_K^{-1}={\epsilon }_K$ to obtain $S_K(t)a={\epsilon }_K(a)S_K(t)$. Conversely, for $r\in {\int }_K^r$, applying $S_K$ to $r{S}_K^{-1}(a)={\epsilon }_K(a)r$ gives $a{S}_K(r)={\epsilon }_K(a)S_K(r)$. The first inclusion makes the right-integral space nonzero and at least one-dimensional; the second inclusion and injectivity make its dimension at most the one-dimensional left-integral space. Hence it is one-dimensional and $S_K({\int }_K^{\mathrm{\ell }})={\int }_K^r$.