For $H$ a Hopf algebra, let $V$ be a right $H$comodule with coaction $\Delta_R$. Moreover, let $W$ be a subspace of $V$ such that $\Delta_R(W) \subseteq W \otimes H$, and note that this implies that $\Delta_R$ restricts to a coaction $V/W \to V/W \otimes H$. If we denote, $$ V^H := \lbrace v \in V ~  ~ \Delta_R(v) = v \otimes 1 \rbrace, $$ and analagously $$ (V/W)^H := \lbrace [v] \in V/W) ~  ~ \Delta_R([v]) = [v] \otimes 1 \rbrace, $$ where $[v]$ denotes the coset of $v$, and $\pi:V \to V/W$ be the canonical projection, then when do we have $$ \pi(V^H) = (V/W)^H? $$
1 Answer
Let $\newcommand\Com{\mathsf{Com}^H}\newcommand\Vect{\mathsf{Vect}}\Com$ be the category of right $H$comodules. This has sufficiently many injectives, so we can compute the right derived functors of the left exact functor $F=\hom_{\Com}(k,\mathord):\Com\to\Vect$, where $K$ denotes the trivial, $1$dimensiona comodule. We can write $\newcommand\Ext{\mathrm{Ext}}\Ext_{\Com}^p(k,V)=R^pF(V)$.
Notice that $F(V)=V^H$ for all $V\in\Com$.
If $$\tag{$\star$}0\to W\to V\to U\to0$$ is a short exact sequence in $\Com$, then we have a long exact sequence for the derived functors $R^pF$, which starts with $$0\to W^H\to V^H\to U^H\to\Ext^1_{\Com}(k,W)\to\cdots$$
We can conclude, then, as usual, that the map $V^H\to U^H$ is surjective if, for example $\Ext^1_{\Com}(k,W)=0$. This can happen for various reasons: one obvious one is that $W$ be an injective comodule. A draconiant version of this is the condition that $H$ be cosemisimple.
To say something more intelligent, one would probably need to know more details about your concrete situation, though.
Dually, we can use the functor $\newcommand\box{\mathbin{\Box^H}}G=(\mathord)\box k$, the cotensor product with the trivial module $k$. Here the traditional notation for its derived functors is $\newcommand\Cotor{\mathrm{Cotor}^H}\Cotor_p(\mathord,k)=R^pG(\mathord)$. The long exact sequence for the derived functors of $G$ applied to the short exact sequence $(\star)$ is now $$0\to W^H\to V^H\to U^H\to\Cotor_1(W,k)\to\cdots$$ and we see that for $V^H\to U^H$ to be surjective is enough that $W$ be coflat.

$\begingroup$ Thanks for your answer. I guess it's what I was looking for, since my real question was: Is it obvious that this is always true? You've certainly shown that it isn't. $\endgroup$ Apr 24, 2012 at 21:29