In a CCC with the initial object, can I have a morphism A -> 0? I'm in a cartesian-closed category with the initial object $0$. What are the consequences of having a morphism from some object $A$ to $0$? Does it mean that there is also a global element of $0$, that is $1\to 0$ ?  Does it mean that we have a zero object, $1 \cong 0$ ? Does it mean that all objects are isomorphic?
Assume we don't have coproducts.

Since the asker is not satisfied with Qiaochu's answer I'll try to expand on it and show more of the intermediate steps.
First we show that $X \times 0 \cong 0$, by showing that $X \times 0$ is an initial object.
$$Hom(X \times 0, A) \cong Hom(0, A^X) \implies |Hom(X \times 0, A)| = 1$$
This means that the projection $\pi_0: X \times 0 \to 0$ is an isomorphism. We use the $Hom(A, -)$ functor to get the isomorphism (functors send isomorphisms to isomorphisms).
$$Hom(A, \pi_0) : Hom(A, X \times 0) \to Hom(A, 0)$$
which is just post-composition by $\pi_0$.
The universal property of the product gives us another isomorphism, which we'll call $f$:
$$ f: Hom(A, X)\times Hom(A,0) \to Hom(A, X \times 0) $$
We can compose these two isomorphisms to get a third isomorphism:
$$ Hom(A, \pi_0) \circ f : Hom(A, X)\times Hom(A,0) \to Hom(A, 0)$$
It turns out that $ Hom(A, \pi_0) \circ f = \pi_{Hom(A,0)} $. This is because
$$ (Hom(A, \pi_0) \circ f)(g,h) = \pi_0 \circ f(g,h) = h $$
The last equality is due to the universal property of $A \times 0$. Hence the projection:
$$ \pi_{Hom(A,0)}:  Hom(A, X)\times Hom(A,0) \to Hom(A, 0)$$
is an isomorphism. If $Hom(A, 0)$ is non-empty then $Hom(A, X)$ also has to be non-empty. If $Hom(A, X)$ had more than one element then $ \pi_{Hom(A,0)} $ would not be injective which means $ Hom(A, X)$ has exactly one element and $A$ is an initial object.
