TITLE: Hom between Brody hyperbolic varieties
QUESTION [5 upvotes]: If $X$ and $Y$ are smooth projective Brody hyperbolic varieties is $\mathrm{Hom}(X, Y)$ also Brody hyperbolic?

REPLY [7 votes]: I assume that for $\operatorname{Hom}(X,Y)$ you mean $\operatorname{Hol}(X,Y)$, that is the family of all holomorphic maps from $X$ to $Y$, endowed with its universal complex structure (which exists since your $X$ is compact).
As you probably know, for a compact complex space being Kobayashi hyperbolic is equivalent to being Brody hyperbolic.
Now, Theorem (6.4.1) in S. Kobayashi "Hyperbolic complex spaces" gives you the answer. Among other things it is stated there what follows.
Even if you assume $X,Y$ to be merely compact complex spaces (no need of any projectivity assumption), and only $Y$ to be hyperbolic, then $\operatorname{Hol}(X,Y)$ is compact and any of its connected component is compact hyperbolic (and hence Brody hyperbolic).
Aside comment (related to Piotr Achinger): if you look at the space of surjective holomorphic maps (or, more generally, dominant meromorphic maps), then under the same assumption as above, it is finite (Theorem (6.6.2) op. cit.).