Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f\colon X \rightarrow Y$ be a morphism of schemes. Let $g\colon Y' \rightarrow Y$ be a morphism. Let $X' = X\times_Y Y'$. Let $f'\colon X' \rightarrow Y'$ be the projection. We are interested in the relation between the set theoretic image of $f'$ and that of $f$. Namely $f'(X') = g^{-1}(f(X))$?

share|improve this question

2 Answers 2

up vote 2 down vote accepted

Yes. The containment $f^\prime(X^\prime)\subseteq g^{-1}(f(X))$ is immediate from commutativity of the relevant cartesian square. Suppose conversely that $y^\prime\in g^{-1}(f(X))$. So $g(y^\prime)=f(x)$ for some $x\in X$. Since $y^\prime$ and $x$ map to the same place in $Y$, there is $x^\prime\in X^\prime$ with $f^\prime(x^\prime)=y^\prime$ and $g^\prime(x^\prime)=x$. In particular, $y^\prime\in f^\prime(X^\prime)$.

The existence of $x^\prime$ follows (basically) from Matt E's answer to your question Set theoretic image of the structure morphism of a fiber product of schemes.

Explicitly, to get $x^\prime$, choose a maximal ideal $\mathfrak{m}\in\mathrm{Spec}(k(x)\otimes_{k(y)}k(y^\prime))$, where $f(x)=y=g(y^\prime)$. Then the composite $\mathrm{Spec}((k(x)\otimes_{k(y)}k(y^\prime))/\mathfrak{m})\rightarrow\mathrm{Spec}(k(x)\otimes_{k(y)}k(y^\prime))\rightarrow X\times_YY^\prime$ sends the unique point of the source to a point $x^\prime$ with the desired projections to $X$ and $Y^\prime$.

share|improve this answer

Look at the universal property of fiber products, and consider maps of a single point into your space (let's assume there's a fixed base field we're working over). If a point maps into both $X$ and $Y'$, and if both projectns of that point agree in $Y$, then we get a unique map of that point into the fiber product. On the other hand, points in the fiber product project to points in $X$ and $Y'$ that both project to the same point in $Y$. Thus, the points in the fiber product are pairs of points in $X$ and $Y'$ that have the same image in $Y$, so your guess is correct.

(Note: I'm not an expert, so anyone can feel free to disagree).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.