Generators of the Chow ring of a space with a cellular decomposition I am reading a paper "Algebraic varieties with rational dissections" by Chow. Here is a link: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC528229/?page=3 . 
A variety $V$ is said to have a rational dissection (or cellular decomposition) if $V$ is a disjoint union of varieties $\{W_i\}$ and each $W_i$ is isomorphic to an affine space (or more generally birational). These $W_i$'s are called cells, or $r$-cells if they have dimension $r$. 
Chow proves the following lemma: Let $W$ be an $r$-cell in a rational dissection of $V$, let $B = \bar{W} - W$ be the boundary of $B$. Then any $s$-cycle $Z$ ($s < r)$ in $V$ with support in $\bar{W}$ is rationally equivalent on $\bar{W}$ to an $s$-cycle $Z_0$ whose support is contained in $B$. 
Now let $W_1,...,W_t$ be the set of all $s$-cells in $V$. Chow claims that by repeated application of the lemma above it follows that every $s$-cycle in $V$ is rationally equivalent to an $s$-cycle whose support is contained in $\bigcup_{i=1}^t \bar{W}_i$, i.e., every cycle is rationally equivalent to a linear combination of the $\bar{W}_i$.  
My problem is that I don't see how we can apply the lemma. The lemma seems to concern lower dimensional cycles, so I don't see how it can be used to prove the claim. Could someone explain this implication in very elementary terms?
 A: Let's back off and think about a more simple situation. Suppose you have a topological space, $B$ and a map $f:S^{r-1}\rightarrow B$ where $S^{d-1}$ is the sphere, thought of as the boundary of of the $r$-ball, which I must unfortunately call $D$, having used up $B$.  
Let $X=B\cup_f D$. We have a Mayer-Vietoris sequence coming from decomposing $X$ as $U=X-\{pt\}$, $V=int(D)$.  With $U\cap V$ homotopy equivalent to an $r-1$ sphere.  Lets assume that $r>1$ just to keep the intersection connected.  
For $s<r-1$ the Mayer-Vietoris sequence yields $H_s(U)\oplus H_{r-1}(V)\rightarrow H_s(X)$ is an isomorphism. However $U$ has $B$ as a strong deformation retract, and $H_{r-1}(V)=0$, which means
$H_s(B)\rightarrow H_s(X)$ is an isomorphism. Any cycle in $X$ is homologous to a cycle in $B$. 
Let's look at the last step,
$$ H_r(X)\rightarrow H_{r-1}(S^{r-1})\rightarrow H_{r-1}(U)\oplus H_{r-1}
(V)\rightarrow H_{r-1}(X)->0$$
Since $H_{r-1}(V)=0$, as long as $H_r(X)\rightarrow H_{r-1}(S^{r-1})$ is onto, you get the desired result there. In fact by induction you know the space you are attaching to has no homology in odd dimensions, and you are only adding even dimensional cells.
I haven't read the paper, but my guess is that if this isn't the argument something like it is.  And now you just add the cells one at a time with increasing dimension.
Yeah, Yeah, I was arguing about singular homology on topological spaces, and he is arguing about the Chow ring, but they are analogous, and chances are there is a way of getting what I said to work out in his setting. :)
