Injectivity of the push-forward of cycles modulo rational equivalence Let $Y$ be a closed subscheme of a scheme $X$, and let $i: Y \rightarrow X$ be the inclusion morphism. We have the push-forward $i_{\ast}: A_{k}(Y) \rightarrow A_{k}(X)$, where $A_{k}(X)$ denotes the group of $k$-cycles modulo rational equivalence.
Under what conditions $i_{\ast}$ is injective?
I am interested in the case $X = Gr(k, n)$ and $Y = H \cap Gr(k, n)$, where $H$ is a general hyperplane.
Thank you!
 A: Disclaimer: This is not an answer, just some thoughts that might help rla.
I think (or rather, I expect that) the obstruction to your map $i_\ast$ being injective has to do with "higher Chow groups" of the complement of $Y$ in $X$. I have to admit I don't know precisely what these are, but I know a bit about the "K_0-theoretic point of view", and there's a precise theorem relating graded $K_0$-theory (for coherent sheaves on $X$) with Chow groups modulo torsion making it plausible that things that are true in $K$-theory might be true in Chow theory; that's an imprecise statement which you should not take too seriously. 
In fact, there is a exact sequence called the localization sequence $$K_0(Y)\to K_0(X) \to K_0(X\backslash Y)\to 0,$$ and the same holds for the Chow ring $$A(Y)\to A(X) \to A(X\backslash Y)\to 0.$$ In $K_0$-theory, I've seen that this localization sequence can be continued by plugging in higher $K$-groups. Therefore, I suspect the same can be done in Chow theory. I can look up a reference for these things if you'd like, although I'm not sure whether it'd be a good one.
Anyway, in case $X=\mathbf P^n$ and $Y$ is a hyperplane, I think you can explicitly compute the map $A(Y) \to A(X)$ by using that the Chow ring of projective $n$-space can be computed explicitly. It's something like  $\mathbf Z[x]/(1-x)^{n+1}$ with $x$  the class of $\mathcal{O}(1)$ if I recall correctly. You can probably also compute the Chow ring of the Grassmannian explicitly, and thus the map in your question too.
