The Hom scheme of two projective varieties over some field is constructed as an open subfunctor of the Hilbert scheme of the product of the two schemes by Grothendieck. So it is a countable union of quasi-projective varieties. For example even in the simplest case $\mathcal{Hom}(\mathbb{P}\_k^1, \mathbb{P}\_k^1)$ there seems to be infinitely many components. So my question is threefold:


1. Is there some general condition that ensures it is a finite union of quasi-projective varieties?
2. Is there some general condition that ensures each component in the countable union is a projective variety?
3. For what type of morphisms $f: X\rightarrow Y$ can one say the map $\mathcal{Hom}(Z,X)\rightarrow \mathcal{Hom}(Z,Y)$ is a proper morphism on the connected components? I know that if $f$ is a closed immersion, the induced map is also closed. I was speculating that if $f$ is finite then the induced map is going to be proper, is something like this true?



When Hom scheme has projective components?
Let $k$ be an algebraically closed field of characteristic zero and let $X$ be a projective variety over $k$. To try and answer your first two questions, let us try formalize the property that Hom-schemes $\mathrm{Hom}(Y,X)$ are finite unions of quasi-projective schemes.


**Definition.** We say that $X$ is *bounded over $k$* if, for every normal projective variety $Y$ over $k$, the Hom-scheme $\mathrm{Hom}(Y,X)$ is of finite type over $k$.


We can then show the following result; see [1] and [2].


(I think that your question is only concerned with 1 and 6 in the following result, but the rest might also be useful to you.)


**Theorem.** The following are equivalent.


1. $X$ is bounded over $k$.
2. For every algebraically closed field $L$ containing $k$, the projective variety $X\_L$ is bounded over $L$.
3. For every smooth projective curve $C$ over $k$, the Hom-scheme $Hom(C,X)$ is of finite type.
4. For every normal projective variety $Y$ over $k$, the Hom-scheme $Hom(Y,X)$ has only finitely many connected components.
5. For every normal projective variety $Y$ over $k$, the Hom-scheme $Hom(Y,X)$ is quasi-projective.
6. For every normal projective variety $Y$ over $k$, the Hom-scheme $Hom(Y,X)$ is a projective scheme over $k$.
7. For every normal projective variety $Y$ over $k$, the Hom-scheme $Hom(Y,X)$ is a projective scheme over $k$. The subscheme of non-connected morphisms $Hom^{nc}(Y,X)$ is a **bounded** projective scheme of dimension $<\dim X$ which maps finitely to $X$.


Boundedness is most likely equivalent to "hyperbolicity". In fact, it is implied by hyperbolicity (Kobayashi, Brody, ...). More precisely:



> 
> 
> > 
> > Theorem. Assume $k=\mathbb{C}$. If $X$ is hyperbolic (i.e., every holomorphic map $\mathbb{C}\to X^{an}$ is constant), then $X$ is bounded.
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> 
> 
> > 
> > Conjecture. Assume $k=\mathbb{C}$. If $X$ is bounded over $\mathbb{C}$, then $X$ is hyperbolic.
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> 


This should answer your first question. For your second question: if $X$ has no rational curves, then every Hom-scheme $Hom(Y,X)$ has projective components (but it can have infinitely many components). Conversely, if all Hom-schemes $Hom(Y,X)$ have (only) projective components, then $X$ has no rational curves. (Use that the components of $Hom(\mathbb{P}^1,\mathbb{P}^1)$ are affine varieties of increasing dimension.)


Abelian varieties have no rational curves and give examples of non-finite type Hom-schemes with each component projective. Note that non-trivial abelian varieties are far from being hyperbolic, so there's no contradiction to the above conjecture.


References.


[1] R. van Bommel, A. Javanpeykar, L. Kamenova.
*Boundedness in families with applications to arithmetic hyperbolicity*
<https://arxiv.org/abs/1907.11225>


[2] A. Javanpeykar and L. Kamenova
*Demailly's notion of algebraic hyperbolicity: geometricity, boundedness, moduli of maps*
<https://arxiv.org/abs/1807.03665>


