Reladenine -- your $X$ is the complement in the affine 3-space of the union of two hypersurfaces $Y$ and $Z$, the first given by $x=0$, the second by $yz=x$. The intersection $Y\cap Z$ is the union of two intersecting affine lines. Moreover, both $Y$ and $Z$ are isomorphic to $\mathbb{C}^2$ (note that $Z$ is the graph of a function). So the Borel-Moore homology of $Y\cup Z$ is given by $H^{BM}_i(Y\cup Z)=\mathbb{C}^2$ if $i=4,3$ and $\mathbb{C}$ if $i=2$. So, by the Alexander duality $\tilde H^j(X)\cong H^{BM}_{6-1-j}(Y\cup Z)$ where $\tilde H$ stands for the reduced cohomology, $H^i(X)=\mathbb{C}$ if $i=0,3$ and $\mathbb{C}^2$ if $i=1,2$. Applying the Poincar\'e duality $H^i(X)\cong H^{6-i}_c(X)$ one gets the answer you give in your posting.
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Reladenine -- your $X$ is the complement in the affine 3-space of the union of two hypersurfaces $Y$ and $Z$, the first given by $x=0$, the second by $yz=x$. The intersection $Y\cap Z$ is the union of two intersecting affine lines. Moreover, both $Y$ and $Z$ are isomorphic to $\mathbb{C}^2$ (note that $Z$ is the graph of a function). So the Borel-Moore homology of $Y\cup Z$ is given by $H^{BM}_i(Y\cup Z)=\mathbb{C}^2$ if $i=4,3$ and $\mathbb{C}$ if $i=2$. So, by the Alexander duality, $H^i(X)=\mathbb{C}$ if $i=0,3$ and $\mathbb{C}^2$ if $i=1,2$. Applying the Poincar\'e duality one gets the answer you give in your posting. |
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