What is an example of a pair of Hopf algebras $(A,B)$ with a surjective Hopf algebra map $\phi:A \to B$ such that $\phi$ does not admit a $B$bicomodule splitting $s:B \to A$? To be clear, the right $B$comodule structure on $A$ is given by $$ (\textrm{id} \otimes \phi) \circ \Delta_A: A \to A \otimes B, $$ where $\Delta_A$ is the coproduct of $A$, and the left coaction is defined similarily.

1$\begingroup$ Did you tried with group algebras? (And a surjective, non split, group homomorphism ) $\endgroup$– Marco FarinatiCommented Sep 22, 2020 at 23:26

$\begingroup$ At a glance, it looks like group algebras might actually not yield the example; if you identify the quotient group $Q = G/N$ with a subset $\tilde Q$ of $G$ via an arbitrary lifiting, the inherited coalgebra structure on the lift $k \cdot \tilde Q$ is extended linearly from $\Delta(q) = q \otimes q$, so there's a coalgebraic splitting, just not an algebraic one. Or am I missing something? $\endgroup$– jdcCommented Sep 26, 2020 at 5:36

$\begingroup$ It's unlikely to be what you want, but in case it's not an accident that you haven't specified a fixed base ring, there are examples that fail to admit even grouptheoretic splittings, e.g., $\mathbb Z[x] \to \mathbb F_2[x]$. $\endgroup$– jdcCommented Sep 26, 2020 at 5:42
1 Answer
I'll give an example "occuring in nature." It's not the simplest possible, but you can get a simpler one by removing the generators of degrees 3, 5, and 7, which don't feature in the argument.
According to results of Borel from 1954, the mod2 homology Hopf algebra $$H_9 = H_* (\mathrm{Spin}(9);\mathbb F_2)$$ is the exterior algebra on one generator each of degrees 3, 5, 6, 7, and 15. The standard inclusion $i$ of $\mathrm{Spin}(9)$ in $\mathrm{Spin}(10)$ preserves this exterior algebra structure, but $H_{10} = H_*(\mathrm{Spin}(10);\mathbb F_2)$ has a new generator $u_{9}$ of degree 9 such that $H_{10}$ is a free module of rank two over $i_* H_9$ with basis $\{1,u_9\}$, following from the collapse of the Serre spectral sequence of the fiber bundle $\mathrm{Spin}(9) \to \mathrm{Spin}(10) \to S^9$. The new generator $u_9$ doesn't anticommute with the old ones as one might expect: giving the other generators the obvious names, one has
$$u_6 u_9 + u_9 u_6 = u_{15}$$
in $H_{10}$.
Particularly, the injection of left $H_9$modules $H_9 \to H_{10}$ does not split. For degree reasons, $u_9$ would have to go to $u_3 u_6$ or zero under the splitting, but the product $H_9 \otimes H_{10} \to H_{10}$ sends $$u_6 \otimes u_9 \mapsto u_{15} + u_9 u_6,$$ while the product $H_9 \otimes H_9 \to H_9$ sends $$u_6 \otimes 0 \mapsto 0 \neq u_{15} + 0 u_6$$ and also $$u_6 \otimes u_3 u_6 \mapsto 0 \neq u_{15} + (u_3 u_6) u_6.$$
A $H^*(\mathrm{Spin}(9);\mathbb F_2)$comodule splitting of the cohomological Hopf algebra map $$i^*\colon H^*(\mathrm{Spin}(10);\mathbb F_2) \to H^*(\mathrm{Spin}(9);\mathbb F_2)$$ would lead on dualization to a forbidden module splitting of the sort we ruled out in the previous paragraph, so $i^*$ is an example of the type you wanted.