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Let $H$ be a Hopf algebra, $(V,\Delta_R)$ a right $H$-comodule map, and $f:V \to V$ a right $H$-comodule map. Since by definition we must have, for all $v \in V$, that $$ \Delta_R(f(v)) = \sum f(v_{(0)}) \otimes v_{(1)}, $$ it seems to me that the kernel of $f$ must be a right $H$-sub-comodule. Since for any non-zero $v \in \ker(f)$, $$ 0 = \Delta_R(0) = \Delta_R(f(v)) = \sum f(v_{(0)}) \otimes v_{(1)}. $$ Does this all seem ok?

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By definition of a comodule, we know $$(\Delta_R\circ f)(v) = (f\otimes 1)\circ \Delta_R (v) = \sum_{(v)} f(v_{(0)})\otimes v_{(1)}.$$

To show that $\ker f$ is a subcomodule of $V$, we need to show that $\Delta_R(\ker f) \subset \ker f \otimes H$.

As you correctly noticed for $v\in \ker f$ one gets $(f\otimes 1)\Delta_R(v) = 0$. But over a field $v\otimes w=0$ if and only if $v=0$ or $w=0$. So $\Delta_R(v)$ must in fact lie in $\ker f\otimes H$.

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