A coproduct coideal with nonzero counit does not give a quotient coalgebra map

By [F5], the specified maps extend linearly. On each basis vector $e,f$, coassociativity and both counit identities hold directly from the formulas; linearity then gives those identities on all of $C$. Thus $C$ is a coalgebra by [F1], and both $e$ and $f$ are group-like by [F2].

For $I=ke$, $\mathrm{\Delta }(\lambda e)=\lambda e\otimes e$ is the image of $(\lambda e)\otimes e\in I\otimes C$ under the inclusion-induced tensor map, so it lies in the coproduct-only image sum of [F4]. But $\epsilon (\lambda e)=\lambda$, so $\epsilon (I)=k\mathrm{\ne }0$ by [F6]. The quotient is nonzero because $f\mathrm{\notin }I$, and its projection $q$ satisfies $q(e)=0$.

If any counit $\bar{\epsilon }$ on $C\mathrm{/}I$ made $q$ a coalgebra map, [F1] would give $\bar{\epsilon }(q(e))=\epsilon (e)=1$. But $q(e)=0$, so linearity gives $\bar{\epsilon }(q(e))=0$, contradicting $0\mathrm{\ne }1$ by [F6]. Therefore no quotient coalgebra structure can make the canonical projection a coalgebra map; in particular the formula $\bar{\epsilon }(c+I)=\epsilon (c)$ is not well-defined.