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Stream: theory: category theory

Topic: Fiber object

Jean-Baptiste Vienney (Jan 04 2025 at 22:54):

Has this notion been studied? Is it a special case of a known concept?

Let $\mathcal{C}$ be a category together with a distinguished object $\top \in \mathcal{C}$ (typically the terminal object but in the case of the category of $k$-modules, I would choose $\top :=k$). Let $f:X \rightarrow Y$. We define a fiber object of $f$ as an object $X_f$ together with an epimorphism $\pi_f:X \rightarrow X_f$ (requiring $\pi_f$ to be an epimorphism ensures that $X_f$ is unique up to isomorphism if it exists) such that the following universal property is satisfied:

For every object $Z$ and morphism $u:X \rightarrow Z$ if for any two morphisms $x,y:\top \rightarrow X$ such that $f \circ x=f \circ y$, we have that $u \circ x=u \circ y$, then there exists a unique morphism $\tilde{u}:X_f \rightarrow Z$ such that $u=\tilde{u} \circ \pi_f$.

In most categories of structured sets and homomorphisms, $X_f$ should be isomorphic to the image of $f$. However, I don't see how the definition of fibre object of a morphism could be equivalent to the definition of image of a morphism in general. I rather expect, but I could be wrong, that only under some hypotheses on $\mathcal{C}$, we have can derive a very general first isomorphism theorem which gives that the image of $f$ is isomorphic to the fibre object $X_f$ of $f$.

Mike Shulman (Jan 04 2025 at 22:58):

If you allow $x,y$ to range over all generalized elements of $X$ (which is usually what you'd want to do), then your $X_f$ is the coequalizer of the kernel of $f$, sometimes called the "regular coimage".

Mike Shulman (Jan 04 2025 at 22:59):

That gives a construction of the usual image (i.e. (regular epi, mono) factorization) in any regular category.

Jean-Baptiste Vienney (Jan 04 2025 at 23:00):

Thank you! This is exactly the answer I was looking for.

Mike Shulman (Jan 04 2025 at 23:00):

Where did you get the term "fiber object" for this? It doesn't look like a fiber at all to me.

Jean-Baptiste Vienney (Jan 04 2025 at 23:01):

If $f:X \rightarrow Y$ is a function and $x \in X$, then $f^{-1}(x)$ is the fiber of $f$ at $x$ and $X_f$ is the set of all these fibers. That's why I thought that fiber object is a good name.

Jean-Baptiste Vienney (Jan 04 2025 at 23:02):

Maybe it is rather "the object of all fibers".

Jean-Baptiste Vienney (Jan 05 2025 at 00:04):

It's weird that every regular category has (regular epi, mono) factorization, but that $\mathbf{Top}$ has (regular epi, mono) factorization without being a regular category.

Jean-Baptiste Vienney (Jan 05 2025 at 00:12):

(Regular epis are the quotient maps in $\mathbf{Top}$.)

Jean-Baptiste Vienney (Jan 05 2025 at 00:21):

Ok, now it feels less weird, seeing that:
Screenshot 2025-01-04 at 7.20.48 PM.png
Regular categories are exactly the categories with (regular epi, mono) factorization + an additional property (which is not satisfied by $\mathbf{Top}$).

Mike Shulman (Jan 05 2025 at 17:55):

Normally "fiber object" refers to an analogue of $f^{-1}(y)$, namely the pullback of $f$ along a global element $y:1\to Y$ (of the codomain, not the domain).