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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?

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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.

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    $\begingroup$ Is it because $\text{Hom}_S(T,S)$ consists entirely of the structure morphism of $T$, and hence a singleton, so the isomorphism follows? $\endgroup$
    – MathChopper
    Nov 20, 2019 at 1:00
  • $\begingroup$ Yep, that's exactly it! $\endgroup$
    – RghtHndSd
    Nov 20, 2019 at 1:59

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