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If I'm given a map from $\mathbb{C}P^1\times\mathbb{C}P^1$ to $\mathbb{C}P^3$ which sends $([z_{0},z_{1}],[w_{0},w_{1}])$ to $[z_{0}w_{0},z_{0}w_{1},z_{1}w_{0},z_{1}w_{1}]$, how do I compute the induced map on the second homology group $H_{2}$?

So far I've tried to find a homotopy of this map to a cellular map so I can compute the induced map using cellular homology. I think there is a homotopy taking the map $[z_{0}w_{0},z_{0}w_{1},z_{1}w_{0},z_{1}w_{1}]$ to the map $[z_{0}w_{0},z_{0}w_{1}+z_{1}w_{0},z_{1}w_{1},0]$, but I'm still having trouble computing the induced map on homology from here. I think that each copy of $\mathbb{C}P^1$ can be viewed as $([z_{0},z_{1}],[1,0])$ and $([1,0],[w_{0},w_{1}])$ which would mean the two generators of $H_{2}(\mathbb{C}P^1\times\mathbb{C}P^1)$ are sent to $[z_{0},z_{1},0,0]$ and $[w_{0},w_{1},0,0]$. Does this mean the induced map would send $(1,0), (0,1)\in\mathbb{Z}\times\mathbb{Z}$ to $1\in\mathbb{Z}$?

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In your map, $w_1$ appears 3 times - was that intentional? Also, overall, I think your second approach will work fine. The key point is that $f_\ast:H_2(\mathbb{C}P^1\times\mathbb{C}P^1 \rightarrow H_2(\mathbb{C}P^3)$ is linear, so it's enough to see what it does on each piece (as you did). –  Jason DeVito Dec 19 '12 at 16:56
    
That was actually a typo, though the map is written correctly now. –  Ryan Hunter Dec 19 '12 at 18:18
    
Great. Did the second part of my comment make sense? If so, feel free to write up an answer yourself and post it. –  Jason DeVito Dec 19 '12 at 21:44

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