A counital coideal quotient of the two-by-two comatrix coalgebra

The counit vanishes on $c_{12}$ and $c_{21}$, and $\epsilon (c_{11}-c_{22})=1-1=0$ by [F2]. The subspace $I$ is nonzero because it contains the basis vector $c_{12}$. Write $j_I:I\hookrightarrow C$ for the inclusion. The formulas $\mathrm{\Delta }(c_{12})=c_{11}\otimes c_{12}+c_{12}\otimes c_{22}$ and $\mathrm{\Delta }(c_{21})=c_{21}\otimes c_{11}+c_{22}\otimes c_{21}$ place their terms in $\mathrm{im}({\mathrm{i}\mathrm{d}}_C\otimes j_I)+\mathrm{im}(j_I\otimes {\mathrm{i}\mathrm{d}}_C)$. For the third generator, $\mathrm{\Delta }(c_{11}-c_{22})=(c_{11}-c_{22})\otimes c_{11}+c_{22}\otimes (c_{11}-c_{22})+c_{12}\otimes c_{21}-c_{21}\otimes c_{12}$, and its four terms lie in the same sum of images. By linearity, $I$ satisfies the field-free coideal condition [F3].

Define $\mathrm{\ell }:C\to k$ on the basis by $\mathrm{\ell }(c_{11})=\mathrm{\ell }(c_{22})=1$ and $\mathrm{\ell }(c_{12})=\mathrm{\ell }(c_{21})=0$, extending linearly by [F5]. Its kernel is exactly $I$: if $z=a{c}_{11}+b{c}_{12}+d{c}_{21}+e{c}_{22}$ then $\mathrm{\ell }(z)=a+e=0$ precisely when $e=-a$, leaving $z$ in the span defining $I$. Every class is a scalar multiple of $u=q(c_{11})=q(c_{22})$, while $u\mathrm{\ne }0$ because $c_{11}\mathrm{\notin }I$: otherwise $\mathrm{\ell }(c_{11})=0$, contrary to $\mathrm{\ell }(c_{11})=1$ and [F7]. Also $q(c_{12})=q(c_{21})=0$. Hence $u$ is a basis of $C\mathrm{/}I$.

Define linear maps $\bar{\mathrm{\Delta }}:C\mathrm{/}I\to (C\mathrm{/}I)\otimes (C\mathrm{/}I)$ and $\bar{\epsilon }:C\mathrm{/}I\to k$ on the basis $u$ by $\bar{\mathrm{\Delta }}(u)=u\otimes u$ and $\bar{\epsilon }(u)=1$, extending them linearly by [F5]. On $u$, coassociativity gives $\alpha (\bar{\mathrm{\Delta }}\otimes \mathrm{i}\mathrm{d})\bar{\mathrm{\Delta }}(u)=u\otimes (u\otimes u)=(\mathrm{i}\mathrm{d}\otimes \bar{\mathrm{\Delta }})\bar{\mathrm{\Delta }}(u)$, and the left and right counit composites both give $u$. Since $u$ is a basis and these are linear maps, the coalgebra identities hold on all of $C\mathrm{/}I$. For $c_{11}$, $(q\otimes q)\mathrm{\Delta }(c_{11})=u\otimes u+0\otimes 0=u\otimes u=\bar{\mathrm{\Delta }}q(c_{11})$ and $\epsilon (c_{11})=1=\bar{\epsilon }q(c_{11})$. For $c_{12}$, $(q\otimes q)\mathrm{\Delta }(c_{12})=u\otimes 0+0\otimes u=0=\bar{\mathrm{\Delta }}q(c_{12})$ and $\epsilon (c_{12})=0=\bar{\epsilon }q(c_{12})$. For $c_{21}$, $(q\otimes q)\mathrm{\Delta }(c_{21})=0\otimes u+u\otimes 0=0=\bar{\mathrm{\Delta }}q(c_{21})$ and $\epsilon (c_{21})=0=\bar{\epsilon }q(c_{21})$. For $c_{22}$, $(q\otimes q)\mathrm{\Delta }(c_{22})=0\otimes 0+u\otimes u=u\otimes u=\bar{\mathrm{\Delta }}q(c_{22})$ and $\epsilon (c_{22})=1=\bar{\epsilon }q(c_{22})$. By linearity these equalities hold on all of $C$, so $q$ is a coalgebra map by [F1].