Counital bases of a weak bialgebra

Taking the middle argument to be $1$ in the weak counit identity gives $\epsilon (gh)=\sum \epsilon (g{1}_{(2)})\epsilon (1_{(1)}h)=\epsilon (g{\epsilon }_t(h))$. Substitution into the definition of ${\epsilon }_t$ yields ${\epsilon }_t(gh)={\epsilon }_t(g{\epsilon }_t(h))$. Coalgebra counitality gives ${\epsilon }_t(1)=1$; setting $g=1$ in the displayed identity gives ${\epsilon }_t^2={\epsilon }_t$. The corresponding identities for ${\epsilon }_s$ follow by reversing the tensor factors.

Let $\mathrm{\Delta }(1)=1_{(1)}\otimes 1_{(2)}=1_{(1)}^{\mathrm{\prime }}\otimes 1_{(2)}^{\mathrm{\prime }}$ denote two copies. The weak-unit identity and multiplicativity give $\mathrm{\Delta }({\epsilon }_t(h))=\epsilon (1_{(1)}h)\mathrm{\Delta }(1_{(2)})=\epsilon (1_{(1)}h)1_{(2)}\otimes 1_{(3)}={\epsilon }_t(h)1_{(1)}\otimes 1_{(2)}=1_{(1)}{\epsilon }_t(h)\otimes 1_{(2)}$. Reversing the tensor factors gives $\mathrm{\Delta }({\epsilon }_s(h))=1_{(1)}\otimes {\epsilon }_s(h)1_{(2)}=1_{(1)}\otimes 1_{(2)}{\epsilon }_s(h)$. The weak-unit and weak-counit equations also give ${\epsilon }_t(1_{(1)}h)1_{(2)}=\epsilon (1_{(1)}^{\mathrm{\prime }}{1}_{(1)}h)1_{(2)}^{\mathrm{\prime }}{1}_{(2)}={\epsilon }_t(h)$ and $1_{(1)}\otimes {\epsilon }_t(1_{(2)})=1_{(1)}\otimes \epsilon (1_{(1)}^{\mathrm{\prime }}{1}_{(2)})1_{(2)}^{\mathrm{\prime }}=1_{(1)}\otimes \epsilon (1_{(2)})1_{(3)}=1_{(1)}\otimes 1_{(2)}$. Hence $\mathrm{\Delta }(1)={\epsilon }_s(1_{(1)})\otimes {\epsilon }_t(1_{(2)})\in H_s\otimes H_t$. The same formulas with the primed counital maps show that they restrict to the identity on their respective images. Thus $\mathrm{\Delta }(x)=x{1}_{(1)}\otimes 1_{(2)}=1_{(1)}x\otimes 1_{(2)}$ for $x\in H_t$, and the factor-reversed formulas hold for $x\in H_s$. Finally, the counital identity $x{\epsilon }_t(h)={\epsilon }_t(xh)$ for $x\in H_t$ gives $xy={\epsilon }_t(xb)\in H_t$ when $x={\epsilon }_t(a)$ and $y={\epsilon }_t(b)$; also $1={\epsilon }_t(1)$. The factor-reversed calculation proves that $H_s$ is a unital subalgebra.

Expanding the counital maps and using the two equal weak-unit products gives ${\epsilon }_t(g){\epsilon }_s(h)=\epsilon (1_{(1)}g)1_{(2)}{1}_{(1)}^{\mathrm{\prime }}\epsilon (h{1}_{(2)}^{\mathrm{\prime }})=\epsilon (1_{(1)}g)1_{(1)}^{\mathrm{\prime }}{1}_{(2)}\epsilon (h{1}_{(2)}^{\mathrm{\prime }})={\epsilon }_s(h){\epsilon }_t(g)$, so the two images commute. For $y\in H_s$, $h\in H$, and $x\in H_t$, the product formula in step 1.1 and the identity $x{\epsilon }_t(h)={\epsilon }_t(xh)$ from step 2.1 give ${\epsilon }_t(yh)={\epsilon }_t(y{\epsilon }_t(h))={\epsilon }_t({\epsilon }_t(h)y)={\epsilon }_t(h){\epsilon }_t(y)$; hence ${\epsilon }_t{\mathrm{\mid }}_{H_s}$ reverses products. Direct substitution into the definitions gives ${\epsilon }_t{\epsilon }_s^{\mathrm{\prime }}(h)=\epsilon (1_{(1)}{1}_{(1)}^{\mathrm{\prime }}\epsilon (1_{(2)}^{\mathrm{\prime }}h))1_{(2)}=\epsilon (1_{(1)}h)1_{(2)}={\epsilon }_t(h)$ by the weak-counit identity. Applying the same calculation with multiplication and coproduct reversed gives ${\epsilon }_s^{\mathrm{\prime }}{\epsilon }_t(h)={\epsilon }_s^{\mathrm{\prime }}(h)$. Since both counital maps are the identity on their images, these equalities make ${\epsilon }_t{\mathrm{\mid }}_{H_s}$ and ${\epsilon }_s^{\mathrm{\prime }}{\mathrm{\mid }}_{H_t}$ mutually inverse. Thus ${\epsilon }_s^{\mathrm{\prime }}(x)\in H_s$; it commutes with $z\in H_t$, and ${\epsilon }_t({\epsilon }_s^{\mathrm{\prime }}(x)z)={\epsilon }_t(z{\epsilon }_s^{\mathrm{\prime }}(x))=z{\epsilon }_t({\epsilon }_s^{\mathrm{\prime }}(x))=zx$.

Put $E=\mathrm{\Delta }(1)$. For every $y\in H$, multiplicativity gives $E\mathrm{\Delta }(y)=\mathrm{\Delta }(y)$; hence $\sum {\epsilon }_t(y_{(1)})y_{(2)}=\sum \epsilon (1_{(1)}{y}_{(1)})1_{(2)}{y}_{(2)}=(\epsilon \otimes \mathrm{i}\mathrm{d})(E\mathrm{\Delta }(y))=y$ by coalgebra counitality. For $x\in H_t$, step 2.1 gives $\mathrm{\Delta }(x)=(x\otimes 1)E=E(x\otimes 1)$, so $\mathrm{\Delta }(hx)=\mathrm{\Delta }(h)(x\otimes 1)E=\mathrm{\Delta }(h)E(x\otimes 1)=\sum h_{(1)}x\otimes h_{(2)}$. Applying the preceding identity to $y=hx$ proves $\sum {\epsilon }_t(h_{(1)}x)h_{(2)}=hx$.

Put $e=({\epsilon }_t\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(1)=\sum e^1\otimes e^2$. Step 2.1 puts both tensor legs in $H_t$, and $e^1=\epsilon (1_{(1)}^{\mathrm{\prime }}{1}_{(1)})1_{(2)}^{\mathrm{\prime }}$, $e^2=1_{(2)}$. For $x\in H_t$, substituting these expressions and using the weak-unit identity yields $\sum \epsilon (x{e}^1)e^2=\epsilon (1_{(1)}^{\mathrm{\prime }}{1}_{(1)})\epsilon (x{1}_{(2)}^{\mathrm{\prime }})1_{(2)}=\epsilon (1_{(1)}x)1_{(2)}=x$. The other contraction is $\sum e^1\epsilon (e^{2}x)=\epsilon (1_{(1)}^{\mathrm{\prime }}{1}_{(1)})1_{(2)}^{\mathrm{\prime }}\epsilon (1_{(2)}x)=\epsilon (1_{(1)}^{\mathrm{\prime }}x)1_{(2)}^{\mathrm{\prime }}={\epsilon }_t(x)=x$, again by the weak-unit identity and $x\in H_t$. Taking $x=1$ in the first contraction gives $\sum \epsilon (e^1)e^2=1$; the weak-unit identity also gives $\sum e^1{e}^2=\sum {\epsilon }_t(1_{(1)})1_{(2)}=1$. The pairing $(y,x)\mapsto \epsilon (yx)$ on $H_t$ is nondegenerate: if $\epsilon (yx)=0$ for every $y\in H_t$, then the second contraction gives $x=\sum e^1\epsilon (e^{2}x)=0$. For any $x,y\in H_t$, both contractions $(\epsilon (y\cdot )\otimes \mathrm{i}\mathrm{d})(xe)$ and $(\epsilon (y\cdot )\otimes \mathrm{i}\mathrm{d})(ex)$ equal $yx$ by the first reconstruction identity; nondegeneracy therefore gives $xe=ex$. Finally, $\mathrm{\Delta }(1)\in H_s\otimes H_t$ and ${\epsilon }_s^{\mathrm{\prime }}{\epsilon }_t{\mathrm{\mid }}_{H_s}={\mathrm{i}\mathrm{d}}_{H_s}$ imply $\mathrm{\Delta }(1)=\sum {\epsilon }_s^{\mathrm{\prime }}(e^1)\otimes e^2$.

For $q\in H_s$, step 2.1 gives $\mathrm{\Delta }(q)=E(1\otimes q)$, so $\mathrm{\Delta }(hq)=\mathrm{\Delta }(h)(1\otimes q)$ and coalgebra counitality yields $\sum h_{(1)}\epsilon (h_{(2)}q)=hq$. The weak counit equation with middle argument $1$ gives $\epsilon (h_{(2)}x)=\sum \epsilon (h_{(2)}{1}_{(1)})\epsilon (1_{(2)}x)$. Since $E\in H_s\otimes H_t$ by step 2.1, applying the preceding contraction to its first leg shows $\sum h_{(1)}\epsilon (h_{(2)}x)=\sum h{1}_{(1)}\epsilon (1_{(2)}x)=h{\epsilon }_s^{\mathrm{\prime }}(x)$. Finally step 3.1 proves ${\epsilon }_s^{\mathrm{\prime }}{\epsilon }_t={\epsilon }_s^{\mathrm{\prime }}$ on all of $H$, and $E\mathrm{\Delta }(h)=\mathrm{\Delta }(h)$ gives $\sum {\epsilon }_s^{\mathrm{\prime }}({\epsilon }_t(h_{(2)}x))h_{(1)}=\sum {\epsilon }_s^{\mathrm{\prime }}(h_{(2)}x)h_{(1)}=\sum 1_{(1)}{h}_{(1)}\epsilon (1_{(2)}{h}_{(2)}x)=\sum h_{(1)}\epsilon (h_{(2)}x)=h{\epsilon }_s^{\mathrm{\prime }}(x)$. The order of the factors is essential: the ${\epsilon }_s^{\mathrm{\prime }}$ factor precedes $h_{(1)}$, as required for the right unit map of left modules. No antipode, finite-dimensionality assumption, or commutativity of $H_t$ has been used.