# fiber product of affine varieties is associative?

Let $$X, Y, Z$$ be affine varieties defined over $$k$$ an algebraically closed field. I was wondering how we know that $$(X \times_k Y) \times_k Z = X \times_k (Y \times_k Z)?$$

Actually I'm not even sure if it's supposed to be "=" here.. What should it be? Any explanation for the two parts would be appreciated. Thank you.

• Affine is not important. For affine varieties, this is just the associativity of tensor products of rings. – Mohan May 15 at 21:55
• As for equality versus isomorphism, fiber products of varieties and tensor products of rings are ordinarily defined only up to isomorphism. Even if you managed to define them more explicitly, it would be difficult to get $=$ rather than $\cong$ in the associative law. – Andreas Blass May 15 at 23:41

Let $$X,Y,Z,A,B$$ be objects in a category $$\mathcal{C}$$ with maps $$f:X\to A$$, $$g:Y\to A$$, $$h:Y\to B$$, $$j:Z\to B$$.
Then the fiber products $$(X\times_A Y)\times_B Z$$ and $$X\times_A(Y\times_B Z)$$ both represent the functor from $$\mathcal{C}^{\text{op}}$$ to $$\mathbf{Set}$$ which takes an object $$W$$ to the collection of all triples of morphisms $$(\alpha,\beta,\gamma)$$, where $$\alpha: W\to X$$, $$\beta: W\to Y$$, $$\gamma:W\to Z$$ such that $$f\circ \alpha = g\circ \beta$$ and $$h\circ \beta = j \circ \gamma$$. Since they both represent the same functor, they are naturally isomorphic.
• @Johnny T, I meant that if we regard the two different orders of fiber products as functors in $X,Y,Z,A,$ and $B$ then they are naturally isomorphic, with the natural isomorphism the one that we get from showing that they both represent the functor I described. – jgon May 16 at 11:46