0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Christoph Nov 22 '18 at 11:28
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.