# The image of a proper scheme is closed

Let $X$ be a scheme, $Y$ a proper $X$-scheme and $Z$ a separated $X$-scheme. Let $f : Y \to Z$ be a morphism of $X$-schemes. I would like to prove that the image of $f$ is closed in $Z$.

Here is what I have so far. The morphism $Y \times_X Z \to Z$ is is a closed (topological) map, as $Y$ is proper over $X$. The morphism $f$ can be factored as $Y \to Y \times_X Z \to Z$, where the first morphism $i$ is given by $(\mathrm{id}_Y, f)$. Since the second map is closed, all I need to prove is that the first map has a closed image. This is where I'm stuck: I don't see where the separatedness of $Z$ can be involved, or how to prove the claim at all.

Am I headed in the right direction? How can I end the proof?

-

Your argument is correct, and you're almost done. The morphism $(\mathrm{id}_Y,f):Y\rightarrow Y\times_XZ$ is called the graph of $f$, sometimes denoted $\Gamma_f$. The fact you need is that, for $Z\rightarrow X$ separated, the graph of any $X$-morphism $Y\rightarrow Z$ is a closed immersion. I don't know how to do diagrams on MSE, but the point is that the graph is the base change of the diagonal $Z\rightarrow Z\times_XZ$ along $f\times\mathrm{id}_Z:Y\times_XZ\rightarrow Z\times_XZ$. Since $Z$ is separated over $X$, its diagonal is a closed immersion, so the graph is as well, being a base change of a closed immersion.
This type of argument is used a lot. I think Ravi Vakil (in his notes) calls it the cancelation principle (something like that). A precise statement is: let $P$ be a property of morphisms which is stable under composition and base change and such that closed immersions have $P$. Then for morphisms $f:Y\rightarrow Z$ and $g:Z\rightarrow X$, if $g\circ f$ has $P$ and $g$ is separated, then $f$ has $P$. The proof is just as you've outlined. You factor $f$ into its graph followed by the projection $Y\times_XZ\rightarrow Z$. The projection is a base change of $g\circ f:Y\rightarrow X$, so it has property $P$ since $g\circ f$ does, and the graph is a closed immersion because $g$ is separated, so it too has property $P$. Thus $f$ has property $P$, being a composite of morphisms with property $P$.
What you're proving is the particular case where $P$ is the property of being proper. Another would be where $P$ is the property of being finite. There is also another version of the cancelation principle where closed immersions are replaced by quasi-compact immersions and separatedness by quasi-separatedness, but the argument is exactly the same.