Which morphisms $1\to 1+1$ can exist? Let $\mathcal{C}$ be a category with finite products and coproducts, and let $f: 1 \to 1+1$, where $+$ denotes the coproduct and $1$ is the final object. Can we conclude that $f$ must be equal to one of the injections $\iota_1,\iota_2$?
Does it help if we also know that $\mathcal{C}$ is a bicartesian closed category?
Context: I was wondering if, in a bi-CCC, seen as a model of the $\lambda$-calculus, it is possible to have more "boolean values" than the two standard ones $\sf true,false$.
In all the categories/models I am familiar with this seems to hold. Yet, after failing to prove it, I am starting to think some more hypotheses might be needed.
 A: In the category of sheaves on a space $X$, the morphisms $1 \to 1+1$ correspond to the open-closed subsets of $X$.
If $X$ is a set with the discrete topology, then this is just $\mathsf{Set}^X$. Here you can also see directly that the morphisms $1 \to 1+1$ correspond to the subsets of $X$, because for each $x \in X$ we may decide which of the two coproduct inclusions we choose.
A: As a more elementary example: Consider the category of two-pointed sets, where the objects are sets with two distinguished (but not necessarily distinct) elements $0$ and $1$, and the morphisms are functions $f$ such that $f(0)=0$ and $f(1)=1$.
This has finite products and coproducts: products are Cartesian products and coproducts are disjoint-except-for-$0$-and-$1$-unions. The initial object is $\{0,1\}$; the final object is a singleton set.
Thus the product $\{0,1\}\times\{0,1\}$ is a set with four elements of which two are $0$ and $1$. When creating a morphism $\{0,1\}\times\{0,1\}\to\{0,1\}$, we can choose the images of the two remaining elements freely, giving $4$ morphisms, only two of which are projections.
Now the opposite of this category is a counterexample to your first question. It is not CCC, though: counting elements and morphisms show that exponentials cannot exist in general.
