I need to prove the following:
"Let $U$ and $V$ be subspaces of a coalgebra $(C,\Delta, \epsilon)$. Suppose that $U\subseteq\ker(\epsilon)$. Show that $U\wedge V\subseteq V$ and $V\wedge U\subseteq V$. Show that it is an equality if $U=\ker(\epsilon)$"
I was able to demonstrate the first part, using the property of counit and comoltiplication. But I can't find a way to solve the second sentence. Any hint would be much appreciated! Thank you!
The fact is that over a field $\Bbbk$ it is always true that $V\subseteq \Delta^{-1}(\ker(\varepsilon)\otimes C+C\otimes V)=\ker(\varepsilon)\wedge V$. The idea is the following: pick a basis for $V$, say $\{e_i\mid i\in I_0\}$, and complete it to a basis of $C$, say $\{e_i,e_j'\mid i\in I_0,j\in I_1\}$. Then you can always write $$\Delta(v) = \sum_ic_i\otimes e_i + \sum_j c_j'\otimes e'_j \qquad \text{for all }v\in V.$$ By the counitality condition, $\sum_j \varepsilon(c_j')e_j'=v-\sum_i\varepsilon(c_i)e_i\in V$. However, by construction, $\sum_j \varepsilon(c_j')e_j'\in V^\perp$ and hence $\sum_j \varepsilon(c_j')e_j'=0$. By linear independence, $\varepsilon(c_j')=0$ for all $j\in I_1$. Conclusion: $$\Delta(v) \in C\otimes V+ \ker(\varepsilon)\otimes C$$ for all $v\in V$.