\section{From Hopf 2-algebra to crossed comodule} For a Hopf 2-algebra as in Definition \ref{Hopf 2-algebra}, consider the correction of characters on $H$ \begin{equation} \label{character induced by s} X :=\{ \phi \in Char(H) \, | \,\, \phi(s(b))=\epsilon_{B}(b),\quad \forall b\in B\}. \end{equation} The set of character on $H$ is a group with convolution product as multiplication, i.e. for any two characters $\phi_{1}$ and $\phi_{2}$ on $H$, we define the multiplication to be $(\phi_{1}\ast \phi_{2})(h):=\phi_{1}(\one{h})\phi_{2}(\two{h})$, for $\forall h\in H$. The the product is clearly associative, since $H$ is co-associative. The unit of this group is the co-unit $\epsilon_{H}$. We can see $X$ is a sub-group of $Char(H)$, since for any $b\in B$ and any $\phi_{1}$, $\phi_{2}\in X$, we have $(\phi_{1}\ast\phi_{2})(s(b))=\phi_{1}(\one{s(b)})\phi_{2}(\two{s(b)})=\phi_{1}(s(\one{b}))\phi_{2}(s(\two{b}))=\epsilon_{B}(\one{b})\epsilon_{B}(\two{b})=\epsilon_{B}(b)$. Then we can define a subalgebra of $H$ \begin{equation} \label{sub Hopf algebra induced by X} E_{X} :=\{ h\in H \, | \,\, \phi(h)=0, \quad \forall \phi\in X\}. \end{equation} It is clear that $E_{X}$ is an ideal of $H$. Moreover, it is a co-ideal of $H$, i.e. $\blacktriangle(E_{X})\subseteq E_{X}\ot H+H\ot E_{X}$. And this can be induced by the definition of convolution product of characters. So the quotient algebra $A:=H/E_{X}$ is Hopf algebra, with all the Hopf algebra structure induced by $H$ in a natural fashion. The element in $A$ is denoted by the equivalence class $[h]$. So the coproduct is given by $\Delta_{A}([h])=\one{[h]}\ot\two{[h]}:=[\one{h}]\ot[\two{h}]$, the counit is given by $\epsilon_{A}([h]):=\epsilon_{H}(h)$ and the antipode is given by $S_{A}([h]):=[S_{H}(h)]$. Moreover, $A$ is also a $B$-bimodule with a trivial left $B$ action by the natural way. Indeed we can define $b\triangleright [h]:=[b\triangleright h]=[s(b)h]=[s(b)][h]=\epsilon_{B}(b)[h]$. However, the right action induced by $t$ is not trivial, it is given by $[h]\triangleleft b:=[h\triangleleft b]=[ht(b)]$. Similarly, we can define \begin{equation} \label{character induced by t} Y :=\{ \phi \in Char(H) \, | \,\, \phi(t(b))=\epsilon_{B}(b),\quad \forall b\in B\}. \end{equation} And \begin{equation} \label{sub Hopf algebra induced by Y} F_{Y} :=\{ h\in H \, | \,\, \phi(h)=0, \quad \forall \phi\in Y\}. \end{equation} Similarly, $F_{Y}$ is also an ideal and co-ideal of $H$, and the quotient Hopf algebra $C:=H/F_{Y}$ can be similarly defined. The element of $C$ is denoted by the equivalence class $$. And similarly $H$ has a $B$-bimodule structure induced by the source and target map descent on $F_{Y}$. \begin{proposition} Let $h\in H$, then we have $[\one{h}]\ot<\two{h}>=[\two{h}]\ot <\one{h}>$. \end{proposition} \begin{proof} First we show For any $h\in H$, we have \begin{align}\label{element in A} [h]=[\epsilon_{H}(\uone{h})\utwo{h}] \end{align} and \begin{align}\label{element in C} =<\uone{h}\epsilon_{H}(\utwo{h})>. \end{align} For (\ref{element in A}), we can check that it is well define over the balance tensor product of $B$. That is $[\epsilon_{H}(\uone{h}\triangleleft b)\utwo{h}]=[\epsilon_{H}(\uone{h})b\triangleright\utwo{h}]$ by using $[s(b)]=\epsilon_{B}(b)$. Then we can see \begin{align*} [h]=[s(\epsilon(\uone{h}))\utwo{h}]=[\epsilon_{B}(\epsilon(\uone{h}))\utwo{h}]=[\epsilon_{H}(\uone{h})\utwo{h}], \end{align*} where the second step use $[s(b)]=\epsilon_{B}(b)$. Similarly, we can also show (\ref{element in C}). Now we can see \begin{align*} [\one{h}]\ot<\two{h}>=&[\epsilon_{H}(\uone{\one{h}})\utwo{\one{h}}]\ot <\uone{\two{h}}\epsilon_{H}(\utwo{\two{h}})>\\ =&[\epsilon_{H}(\one{\uone{h}})\one{\utwo{h}}]\ot <\two{\uone{h}}\epsilon_{H}(\two{\utwo{h}})>\\ =&[\one{\utwo{h}}\epsilon_{H}(\two{\utwo{h}})]\ot<\epsilon_{H}(\one{\uone{h}})\two{\uone{h}}>\\ =&[\epsilon_{H}(\one{\utwo{h}})\two{\utwo{h}}]\ot<\one{\uone{h}}\epsilon_{H}(\two{\uone{h}})>\\ =&[\epsilon_{H}(\utwo{\one{h}})\utwo{\two{h}}]\ot<\uone{\one{h}}\epsilon_{H}(\uone{\two{h}})>\\ =&[\epsilon_{H}(\uone{\two{h}})\utwo{\two{h}}]\ot <\uone{\one{h}}\epsilon_{H}(\utwo{\one{h}})>\\ =&[\two{h}]\ot<\one{h}>, \end{align*} where the 3rd and 5th steps use (\ref{commutation property}). \end{proof} \begin{lemma}\label{lem. commutate subalgebra} For any $h\in H$, we have $[\one{h}t(\epsilon(S_{H}(\two{h})))]\ot [\three{h}]=[\two{h}t(\epsilon(S_{H}(\three{h})))]\ot [\one{h}]$. \end{lemma} \begin{proof} First define a map $G: C\to A$ by $G():=[\one{h}t(\epsilon(S_{H}(\two{h})))]$ for any $h\in H$. This is well defined on $C$, since for any $f\in F_{y}$ we have \begin{align*} G()=[\one{f}t(\epsilon(S_{H}(\two{f})))]. \end{align*} Given $\phi\in Char(H)$, we can define $(1_{t(\phi)})^{-1}\in Char(H)$ by $(1_{t(\phi)})^{-1}(h):=\phi(t(\epsilon(S_{H}(h))))$. We can also see $\phi\ast (1_{t(\phi)})^{-1}\in Y$, since \begin{align*} \phi\ast (1_{t(\phi)})^{-1}(t(b))=&\phi(t(\one{b}))\phi(t(S_{B}(\epsilon(t(\two{b})))))=\phi(t(\one{b}))\phi(t(S_{B}(\two{b})))\\ =&\phi(t(\one{b}S_{B}(\two{b})))=\phi(t(\epsilon_{B}(b)))=\epsilon_{B}(b), \end{align*} for any $b\in B$. Thus, for any $\phi\in X$, we have \begin{align*} \phi(\one{f}t(\epsilon(S_{H}(f))))=(\phi\ast (1_{t(\phi)})^{-1})(f)=0. \end{align*} So this lemma is the result of applying $id_{A}\ot G$ on both side of $[\one{h}]\ot<\two{h}>=[\two{h}]\ot<\one{h}>$. \end{proof} There is a Hopf algebra map $\Phi: B\to A$ defined in the following way \begin{align} \Phi(b):=[t(b)], \end{align} for any $b\in B$. This is a bialgebra map, since it is a composition of two bialgebra maps. Moreover, $A$ is also a left $B$-comodule, with the coaction $\delta:A\to A\ot B$ defined by the following: \begin{align} \delta([h]):=\epsilon(\one{h}S_{H}(\three{h}))\ot[\two{h}], \end{align} for any $[h]\in A$. This coaction is well defined, since $\epsilon(\one{f}S_{H}(\three{f}))\ot \two{f}\in B\ot E_{X}$ for any $f\in E_{X}$. Therefore, %(by applying $(\epsilon_{B}\ot \phi)$ on it, where $\phi\in X$). %This step need to be check!!!!! We can see $\delta$ is coaction, since \begin{align*} (\Delta_{B}\ot id_{A})(\delta([h]))=&\Delta_{B}(\epsilon(\one{h}S_{H}(\three{h})))\ot [\two{h}]\\ =&(\epsilon\ot \epsilon)\Big(\one{\big(\one{h}S_{H}(\three{h})\big)}\ot \two{\big(\one{h}S_{H}(\three{h})\big)}\Big)\ot [\two{h}]\\ =&(\epsilon\ot \epsilon)\big(\one{h}S_{H}(\five{h})\ot \two{h}S_{H}(\four{h})\big)\ot [\three{h}]\\ =&\epsilon(\one{h}S_{H}(\five{h}))\ot \epsilon(\two{h}S_{H}(\four{h}))\ot [\three{h}]\\ =&(id_{B}\ot \delta)(\delta([h])), \end{align*}{} for any $[h]\in A$. And $\epsilon_{B}(\epsilon(\one{h}S_{H}(\three{h})))\ot [\two{h}]=[h]$. Moreover, $A$ is a $B$-comodule algebra, since \begin{align*} \delta([h][g])=&\epsilon(\one{(hg)}S_{H}(\three{(hg)}))\ot[\two{(hg)}]\\ =&\epsilon(\one{h}\one{g}S_{H}(\three{g})S_{H}(\three{h}))\ot [\two{h}][\two{g}]\\ =&\delta([h])\delta([g]), \end{align*}{} for any $[h], [g]\in A$. We can also see $A$ is a comodule coalgebra of $B$. Since we have \begin{align*} \eins{[h]}\ot \one{\nul{[h]}}\ot\two{\nul{[h]}}=&\epsilon(\one{h}S_{A}(\four{h}))\ot [\two{h}]\ot [\three{h}]\\ =&\epsilon(\one{h}S_{H}(\three{h})\four{h}S_{H}(\six{h}))\ot [\two{h}]\ot[\five{h}]\\ =&\epsilon(\one{\one{h}}S_{H}({\three{\one{h}}}))\epsilon(\one{\two{h}}S_{H}(\three{\two{h}}))\ot[\two{\one{h}}]\ot [\two{\two{h}}]\\ =&\eins{\one{[h]}}\eins{\two{[h]}}\ot \nul{\one{[h]}}\ot\nul{\two{[h]}}, \end{align*}{} and \begin{align*} \eins{[h]}\ot \epsilon_{A}(\nul{[h]})=\epsilon(\one{h}S_{H}(\three{h}))\epsilon_{A}([\two{h}])=\epsilon(\one{h}S_{H}(\two{h}))=\epsilon_{A}([h]). \end{align*}{} \begin{theorem} Given a Hopf 2-algebra as in Definition \ref{Hopf 2-algebra}, we can construct a crossed comodule. \end{theorem} \begin{proof} We already see $A$ is a left $B$-comodule algebra and left $B$-comodule coalgebra. So we just need to check condition (2) and (3) in Definition \ref{cross comodule}. For (2) we have \begin{align*} \eins{\Phi(b)}\ot \nul{\Phi(b)}=&\eins{[t(b)]}\ot\nul{[t(b)]}\\ =&\epsilon(\one{t(b)}S_{H}(\three{t(b)}))\ot[\two{t(b)}]\\ =&\epsilon(t(\one{b})t(S_{B}(\three{b})))\ot [t(\two{b})]\\ =&\one{b}S_{B}(\three{b})\ot [t(\two{b})], \end{align*}{} where the last step use $\epsilon\circ t=id_{B}$. For (3) we have, \begin{align*} &[\one{h}S_{H}(\three{h})]\ot[\two{h}]\\ =&[\one{h}t(\epsilon(S_{H}(\two{h})))t(\epsilon(\three{h}))t(\epsilon(S_{H}(\five{h})))t(\epsilon(\six{h}))S_{H}(\seven{h})]\ot[\four{h}]\\ =&[\one{h}t(\epsilon(S_{H}(\two{h})))t(\epsilon(\one{\three{h}}))t(\epsilon(S_{H}(\three{\three{h}})))t(\epsilon(\four{h}))S_{H}(\five{h})]\ot[\two{\three{h}}]\\ =&[\two{a}t(\epsilon(S_{H}(\three{h})))t(\epsilon(\one{\one{h}}))t(\epsilon(S_{H}(\three{\one{h}})))t(\epsilon(\four{h}))S_{H}(\five{h})]\ot[\two{\one{h}}]\\ =&[\four{h}t(\epsilon(S_{H}(\five{h})))t(\epsilon(\one{a}))t(\epsilon(S_{H}(\three{h})))t(\epsilon(\six{h}))S_{H}(\seven{h})]\ot[\two{h}]\\ =&[\four{h}t(\epsilon(\one{h}))t(\epsilon(S_{H}(\three{h})))S_{H}(\five{h})]\ot [\two{h}]\\ =&[t(\epsilon(\one{h}))t(\epsilon(S_{H}(\three{h})))]\ot [\two{h}]\\ =&[t(\epsilon(\one{h}S_{H}(\three{h})))]\ot [\two{h}], \end{align*}{} for any $[h]\in A$, where the third step use Lemma \ref{lem. commutate subalgebra}. \end{proof}

On coherent Hopf 2-algebras

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On coherent Hopf 2-algebras

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