Example [lcc-0009]

In \mathsf {Lens}{\mathsf {Set}} \binom {0}{1} + \binom {1}{1} = \binom {0}{2}. To see this, let \binom {A}{X} be a test object, and consider \operatorname {\mathrm {Hom}}(\binom {0}{1}, \binom {A}{X}) \times \operatorname {\mathrm {Hom}}(\binom {1}{1},\binom {A}{X}). If A is empty, the backwards pass is trivial in both cases, and so this just amounts to choosing two points of X. If A is nonempty, it is impossible to define the backwards pass in the first map (or the forwards pass, if X is also empty), and so this product is empty. On the other hand, \operatorname {\mathrm {Hom}}(\binom {0}{2}, \binom {A}{X}) is exactly the same - if A is empty, it is X^2, and if A is nonempty, it is the empty set. Hence this is a coproduct.