Kernels in a category with zero objects: essential uniqueness

An object $$X$$ of a category $$\mathsf{C}$$ is a zero object if it is both initial and terminal, that is, if for any $$Y \in \mathsf{C}$$ there are unique morphisms $$Y\to X$$ and $$X\to Y$$.

We define a kernel of a morphism $$f\colon X\to Y$$ as an equalizer of $$f$$ and $$g$$, where $$g$$ is the unique morphism $$X \to 0 \to Y$$ for a zero object $$0$$.

However, what if a category has more than one zero object? I know that they are uniquely isomorphic, but how does it help? How can we speak of the essentialy uniqueness of a kernel?

It is known that two different objects $$X$$ and $$Y$$ satisfying the same universal properties are isomorphic. However, given a morphism $$f\colon X\to Y$$ and zero objects $$0_a$$ and $$0_b$$, being an equalizer of $$f$$ together with $$X\to 0_a \to Y$$ and being an equalizer of $$f$$ together with $$X \to 0_b \to Y$$ are two different universal properties.

• There's a unique isomorphism between any kernel with respect to $0_a$ and any kernel with respect to $0_b$ making all the diagrams commute. Hence, even varying the zero object, kernels are defined up to unique isomorphism. – Christoph Nov 22 '18 at 11:28

The zero morphism $$X\to Y$$ is already unique: There are unique arrow $$0_a\to 0_b$$ and $$0_b\to 0_a$$, inverses of each other.
Also, by uniqueness $$X\to 0_b\ =\ X\to 0_a\to 0_b$$ and $$0_b\to Y=0_b\to 0_a\to Y$$.
Putting these together, we see that the two compositions $$X\to 0_a\to Y$$ and $$X\to 0_b\to Y$$ are equal.