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A recent project has forced me to consider a rather special object in a rather nasty category. Consider any category $\mathcal{C}$ which has

  • image objects, meaning for each morphism $f: x \to y$ there exists an object $\text{im }f$ along with morphisms $$x \stackrel{s}{\twoheadrightarrow}\text{im }f \stackrel{i}{\hookrightarrow} y$$ so that $s$ is surjective, $i$ is injective and $f = i\circ s$, plus the obvious universal property,

  • an initial object $\iota$ so that each $x \in \mathcal{C}_0$ admits a unique morphism $\iota \to x$, and

  • a final object $\phi$ so that each $x \in \mathcal{C}_0$ admits a unique morphism $x \to \phi$.

Note that in my case $\iota \neq \phi$, so there is no zero object in $\mathcal{C}$. What we do have instead, is a unique morphism $\iota \to \phi$. This morphism has an associated image object, which I will label $d$.

Has $d$ been defined and studied somewhere? What are its fundamental properties?

I'm sorry if the question is seen as an obvious reference request. Category theory seems particularly rich in this Rumplestiltskin phenomenon, where the difference between finding what you are looking for and flailing around miserably lies simply in knowing a special name.

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When you say "surjective" do you mean "epimorphism" or "regular epimorphism"? How about "injective"? –  Neil Strickland Jul 3 '13 at 20:12
@NeilStrickland, by (in, sur)jective I mean only that the relevant (left, right) cancellation property holds. –  Vidit Nanda Jul 3 '13 at 20:16
Which of the two obvious universal properties is the obvious universal property to which you refer? –  Steven Landsburg Jul 3 '13 at 20:19
@StevenLandsburg sorry, I should have specified: given any object $e$ with surjective $x \to e$ and injective $e \to y$ whose composition equals $f$, there exists a morphism $\text{im }f \to e$ (not the other way around) making things commute. –  Vidit Nanda Jul 3 '13 at 20:22
A more usual definition of $im(f)$ is the smallest subobject through which $f: A \to B$ factors. If the category has equalizers, it may be shown that the map from $A$ to (the domain of) $im(f)$ is an epimorphism. –  Todd Trimble Jul 3 '13 at 20:34

1 Answer 1

d is the initial object $\iota$

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quest, I don't see why this should be the case. In fact, in the category which I'm working with, this is decidedly not the case. –  Vidit Nanda Jul 3 '13 at 20:03
Assume you have $\iota\rightarrow d\rightarrow \varphi$, a decomposition of $f$. Then it factors through $\iota\rightarrow \iota\rightarrow \varphi$ obviously. –  user36465 Jul 3 '13 at 20:14
The assumptions are self-dual, but quest's conclusion is not, so it cannot be correct. –  Neil Strickland Jul 3 '13 at 20:18
In the category of sets under a set $A$ with more than one element, the image is the terminal object, not the initial object. The problem with quest's argument is that the map from the initial object to the terminal object need not be injective. –  Eric Wofsey Jul 3 '13 at 20:21
@quest: that is simply not true. Consider the category of commutative rings with identity, where the initial object is the ring of integers. –  Todd Trimble Jul 3 '13 at 20:40

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