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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 $f'(X') = g^{-1}(f(X))$?

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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 $Y^\prime$.

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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).

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