# Why is the fiber product $X \times_s S$ isomorphic to $X$ itself?

This might be a silly question but I am new to do this and would be appreciated for your help.

Suppose $$X$$ is an $$S$$-scheme, say we have structure morphisms $$f:X \rightarrow S$$ and $$\text{Id}:S\rightarrow S$$. Why is $$X \times_s S \cong X$$?

The definition of fibered product I am using is: A triple $$(Z,p,q)$$ where $$h:Z \rightarrow S$$ is an $$S$$-scheme and morphisms of $$S$$-schemes $$p:Z\rightarrow X$$ and $$q:Z\rightarrow Y$$ is called a fiber product if for every $$S$$-scheme $$T$$, a mapping of sets $$\text{Hom}_S(T,Z)\rightarrow \text{Hom}_S(T,X) \times \text{Hom}_S(T,Y)$$ is bijective.

So in this case I need to show $$\text{Hom}_S(T,X)\rightarrow \text{Hom}_S(T,X) \times \text{Hom}_S(T,S)$$ is a bijection, but why is this true?

Focus on $$\text{Hom}_S(T, S)$$. What is this set? Remember that an element must be an $$S$$-morphism! Drawing out the diagram may help.
One more hint: the structure map of the $$S$$-scheme $$S$$ is the identity map.
• Is it because $\text{Hom}_S(T,S)$ consists entirely of the structure morphism of $T$, and hence a singleton, so the isomorphism follows? Nov 20, 2019 at 1:00