# Cartesian product of two CW-complexes

Let's $A$ and $B$ are CW-complexes. How to construct CW-complex $A\times B$?

Thanks.

## 1 Answer

Choose a CW-complex structure for $A$ using cells $e_\alpha$ with attaching maps $\varphi_\alpha$. Do the same for $B$ using cells $e_\beta$ and attaching maps $\varphi_\beta$. Then the products $e_\alpha \times e_\beta$ are cells and the maps $\varphi_\alpha \times \varphi_\beta$ are attaching maps for a CW-complex structure on $A \times B$.

If you are looking for details of a proof, one can be found in Hatcher's book where the statement appears as Theorem A.6.

• It's worth noting that the topology on such a CW structure for $A\times B$ need not be the product topology. In almost every case we care about it agrees, but one can construct weird examples where the topology is different. Hatcher has such an example in his appendix. – SL2 Jan 18 '12 at 17:31
• @wckronholm: but how to present $S^p\times S^q$ as CW complex using cells of dimension 0, p, q, p+q? – Aspirin Jan 18 '12 at 17:43
• @Aspirin $S^p$ has a CW-complex structure with one $p$-cell $e^p$ and one $0$-cell $e^0$, and $S^q$ has a CW-complex structure with one $q$-cell $e^q$ and one $0$-cell which I will also call $e^0$. The method I mention above then tells you exactly the cells in a CW complex structure for $S^p \times S^q$, namely the cells are $e^p \times e^q$, $e^p \times e^0$, $e^0 \times e^q$, and $e^0 \times e^0$. – wckronholm Jan 18 '12 at 23:15
• A caveat: note that what wckronholm calls "attaching maps" is not what Hatcher calls like that: in Hatcher's nomenclature, these would be the "characteristic maps". Note indeed that the product of disks is a disk, but the product of spheres is not a sphere. – Bruno Stonek Sep 29 '15 at 13:22