Isomorphism between an algebraic set and a cartesian product

Question

Let W, X and Y be algebraic sets and $\gamma_1$: W $\rightarrow$ X, $\gamma_2$: W $\rightarrow$ Y, two morphisms which verify that, given Z an algebraic set and two morphisms $\alpha$: Z $\rightarrow$ X and $\beta$: Z $\rightarrow$ Y, exists only one morphism $\phi$: Z $\rightarrow$ W verifying $\alpha = \gamma_1\phi$ and $\beta = \gamma_2\phi$. Prove that W is isomorphic to X x Y.

Answer 1

Well, you have to show that $X\times Y$ satisfies the same condition as $Z$. Doing this, you will have that there is a unique morphism $X\times Y\to Z$ and a unique morphism $Z\to X\times Y$ (such that the respective diagrams commute), since there is a unique morphism such that makes the diagram commutes and such that $Z\to Z$ is the identity (same argument for $X\times Y\to X\times Y$). You will have that $Z$ is isomorphic (with unique isomorphism) to $X\times Y$.

Now, forget the algebraic structure for a second. $Z$ satisfies the product universal property for sets, so that $Z$ is isomorphic to $X\times Y$ (see the above argument) with a unique isomorphism. Now, if $\phi:X\times Y\to Z$ is the biyection (iso of sets) you have to prove that it is an isomorphism of algebraic varieties. This is, if $\psi:Z\to X\times Y$ if the (set) inverse of $\phi$: Probe that $\phi$ and $\psi$ are morphisms. But this will follow from the diagrams and using the definition of algebraic set morphism.