Let $H=H^0\oplus H^1$ be some super vector space and $E\subseteq H$ a vector subspace. One might think that by setting $$E^k:=E\cap H^k$$ we obtain a new super vector space $E=E^0\oplus E^1$, but this is generally not the case, is it? I don't see why the function $$E^0\times E^1\ni(x,y)\mapsto x+y\in E$$ should be surjective in general. So I think that we need to additionally require that $(x,y)\in H^0\times H^1$ and $x+y\in E$ together implies that $x,y\in E$, right?
Motivation: I think that this relevant for the definition of the index and the proof of the McKean Singer formula, where we consider the super-dimension of the eigenspaces of the Laplacian. So the eigenspaces are viewed as super spaces and as explained above we can not take it for granted that they inherit a grading from the domain $H$ of the Laplacian. But using the fact that the Laplacian is even we can show that the requirement above is satisfied.
This is correct. Maybe it is a good idea to think about it as an equivalent to the fact that submodules of a free module are not necessarily free.