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Let $X,Y,S$ are schemes, and $f:X\to S, g:Y \to S$ are morphisms, does fiber product $X\times_S Y$ always exist in the usual sense(for example as defined in Hartshorne )?

Here is an interesting thing. If $f(X) \cap g(Y)=\emptyset$, there is no way to have their product. Say letting $X=S=Spec(k[x]_x), Y=Spec(k)$, and $g$ corresponding to the ring map $f(x)\in k[x] \to f(0)\in k$, geometrically, $g$ is the embedding of a point to $0 \in \mathbb{A}^1=S$, and $f$ is the open immersion $\mathbb{A}^1-0 \to \mathbb{A}^1$, they satisfy above property.

Do we need to add some restriction to make fiber product always exists?

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    So you do not accept spectrum of the trivial ring?2011-09-13
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    To be honest, I never heard of that. Do you mean in that case, the product should be spectrum of the trivial ring? How do you define the scheme structure on it as well as morphism? Is product unique in that case?2011-09-13
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    1. Yes. By the usual tensor product calculation.2011-09-13
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    2. Empty topological space, and the section over empty set is 0. It would be the initial object in the category of schemes, and I don't think any nonempty scheme has a map to it.2011-09-13
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    3. Product is always unique up to isomorphism by universal property. As for existence, honestly I didn't think about gluing over empty set when I read the proof, but it should work.2011-09-13
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    Everything Soarer said is correct.2011-09-13
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    What do [logic] has to do with all that?2011-09-13
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    As you have seen,here one need the map from empty set to a set, and I guess this depend on what set theory you choose. I just have a vague intuition, it has some relation with the fundations.2011-09-13
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    @Soarer Thank you for your answer, I feel more comfortable with products now.2011-09-13

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