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.
 A: In any cartesian closed category, product $X \times (-)$ distributes over colimits, and in particular over the initial object, so we have $X \times 0 = 0$. If $A$ is any object it follows that
$$\text{Hom}(A, X) \times \text{Hom}(A, 0) \cong \text{Hom}(A, 0)$$
in the sense that the projection map from the LHS to the RHS is an isomorphism, hence that if $\text{Hom}(A, 0)$ is non-empty then $\text{Hom}(A, X) = 1$ for all $X$, so $A = 0$. (Here I am using "equality" to mean that there is a unique map in one direction or another and that unique map is an isomorphism.)
Edit: In more detail, since this was apparently confusing. By $X \times 0 = 0$ I mean that the unique map $0 \to X \times 0$ is an isomorphism. Since this implies $X \times 0$ is the initial object, there is also a unique map $X \times 0 \to 0$, which must be the projection $\pi_0$, and which must be an isomorphism. Now we have the general observation that in any category with finite products, if
$$\pi_Q : P \times Q \to Q$$
is the projection then the induced map
$$\text{Hom}(R, P \times Q) \cong \text{Hom}(R, P) \times \text{Hom}(R, Q) \to \text{Hom}(R, Q)$$
is the projection $\pi_{\text{Hom}(R, Q)}$. So, from the fact that the projection $\pi_0: X \times 0 \to 0$ is an isomorphism, we deduce that for any pair of objects $A, X$ the projection
$$\pi_{\text{Hom}(A, 0)} : \text{Hom}(A, X) \times \text{Hom}(A, 0) \to \text{Hom}(A, 0)$$
is an isomorphism. If $\text{Hom}(A, 0)$ is empty then both sides are empty and we don't learn anything. But if $\text{Hom}(A, 0)$ is non-empty, or equivalently if $A$ admits a morphism to $0$, then the fibers of this map can all be identified with $\text{Hom}(A, X)$, so it follows that if $\text{Hom}(A, 0)$ is non-empty then $\text{Hom}(A, X) = 1$ for all $X$, so $A = 0$.
I had forgotten this argument but apparently I knew it in 2015, although in that answer I wrote it in the opposite category. (We don't use the full strength of the assumption of cartesian closure, only that finite products exist and distribute over initial objects, so this argument continues to work in a distributive category.)
If $A = 1$, or more generally if $A$ has a global point, it follows that the category has a zero object, and now the isomorphisms
$$\text{Hom}(X, Y) \cong \text{Hom}(1, Y^X) \cong \text{Hom}(0, Y^X) \cong 1$$
show that there is a unique morphism between any pair of objects, so the category is contractible.
A: 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.
