The usual way of defining images in category theory is:

Definition 0. Let $f : X \rightarrow Y$ denote a morphism. Then a monomorphism $m : J \rightarrow Y$ is an image of $f$ iff $m$ is initial when viewed as an object of the following category:

Objects. Monomorphisms $m' : J' \rightarrow Y$ into $Y$ such that there exists $e' : X \rightarrow J'$ satisfying $f = m \circ e$.

Arrows. Commutative triangles.

But there seems to be another approach:

Definition 1. Let $f : X \rightarrow Y$ denote a morphism. Then an arrow $m : M \rightarrow Y$ is an image of $f$ iff $m$ is terminal when viewed as an object of the following category.

Objects. Morphisms $m'$ into $Y$ such that for all objects $Z$ and all morphisms $\alpha,\beta : Y \rightarrow Z$, we have $$\alpha \circ f = \beta \circ f \rightarrow \alpha \circ m' = \beta \circ m'.$$

Arrows. Commutative triangles.

Thinking in terms of $\mathbf{Set}$, the first definition emphasizes that $f : X \rightarrow Y$ can be restricted to $f : X \rightarrow \mathrm{img}(f)$, while the second definition (kind of) emphasizes that if we're trying to work out whether or not $\alpha \circ f$ equals $\beta \circ f$, we can focus our attention on whether or not $\alpha \restriction_{\mathrm{img}(f)}$ equals $\beta \restriction_{\mathrm{img}(f)}.$

I have a long list of questions about this:

Question 0. Is the $m$ in Definition 1 always monic?

Question 1. Does $m$ in Definition 1 always satisfy $\alpha \circ m = \beta \circ m \rightarrow \alpha \circ f = \beta \circ f$?

(This is what we need for the set-theoretic intuition about $\restriction_{\mathrm{img}(f)}$ to work, of course.)

Question 2. Does either notion of image imply the other?

  • $\begingroup$ Your second version doesn't quite make sense. In a pointed category, it seems $0\to Y$ would always be the image. Better to ask for $\alpha f=\beta f\equiv \alpha m'=\beta m'$, I would think. It also seems a bit unnatural not to ask explicitly that $f$ factors through the image...up to these caveats, it seems you're not far here from the notion of coimage, the terminal epimorphism out of $X$ factoring $f$. $\endgroup$ – Kevin Carlson Sep 26 '16 at 1:21
  • $\begingroup$ @KevinCarlson, $0 \rightarrow Y$ isn't the image according to definition 1. I think you're reading the word "initial" where I've got the word terminal. $\endgroup$ – goblin Sep 26 '16 at 2:56
  • $\begingroup$ Although come to think of it, we can get a variant on Definition 1 by replacing $\alpha \circ f = \beta \circ f \rightarrow \alpha \circ m' = \beta \circ m'$ with $\alpha \circ m = \beta \circ m \rightarrow \alpha \circ f = \beta \circ f$ and replacing "terminal" with "initial". $\endgroup$ – goblin Sep 26 '16 at 3:02
  • $\begingroup$ @KevinCarlson a map which factors through 0 is the zero map, so that 0 is its image. The factorization has to be equal to $f$ $\endgroup$ – Noix07 Feb 7 '17 at 22:25
  • $\begingroup$ @user39158, yes, you're right. I've fixed the problem. $\endgroup$ – goblin Feb 7 '17 at 23:58

Question 0. Yes, suppose that $m:S\to Y$ is the image of $f:X\to Y$ in the sense of Definition 1 and that for some $u,v:W\to S$, $mu=mv$. But then if $\alpha f =\beta f$ we have by assumption $\alpha m =\beta m$ and hence $\alpha mu = \beta mu$. Therefore $mu:W\to Y$ is an object in the category in which $m$ is terminal. This means that $u=v$ since it is the unique morphism from $mu$ to $m$ in this category.

Question 1. Yes, $f$ is an object in the category in which $m$ is terminal. This means that $f = me$ and hence $\alpha m =\beta m$ implies $\alpha f = \alpha m e = \beta m e =\beta f$.

Question 2. Sometimes. Suppose that $m_0:S_0\to Y$ and $m_1:S_1\to Y$ are images of $f:X\to Y$ in the sense of Definition 0 and Definition 1 respectively. We have a unique morphism $u: S_0 \to S_1$ such that $m_0 =m_1u$ since as explained in Questions 0 and 1 the morphism $m_1$ is a mono and there is a morphism $e_1 : X\to S_1$ such that $m_1e_1=f$. Note that $u$ is necessarily a monomorphism (since $m_0$ is). In general $u$ need not be an isomorphism. For example consider the inclusion $i:\mathbb{Z}\to \mathbb{Q}$ of the integers into the rational numbers in the category of rings. The image in the sense of Definition 0 is $i$ since $i$ is a monomorphism while the image in the sense of Definition 1 is $1_{\mathbb{Q}}$ since $i$ is an epimorphism.

It seems worth mentioning that for a morphism $f:X\to Y$ the image in the sense of Definition 1 can be constructed as the equaliser of the cokernel pair of $f$. Let us see why. Suppose $c_1,c_2: Y\to C$ is the cokernel pair of $f$ and $m:S\to Y$ is the equaliser of $c_1$ and $c_2$. Notice that for a pair of morphisms $\alpha,\beta: Y\to Z$, $\alpha f=\beta f$ if and only if there exists a unique morphism $u: C\to Z$ such that $\alpha = uc_1$ and $\beta = uc_2$. It follows that $m' : S'\to Y$ is in the category described in Definition 1 if and only if $c_1m'=c_2m'$. This proves the claim.

  • $\begingroup$ @user39158 Thanks, I fixed it. $\endgroup$ – Nex Feb 8 '17 at 19:02
  • $\begingroup$ I wonder if I should ask an independent question: but can one explicitly give some conditions on the category so that the two definitions coincide. $\endgroup$ – Noix07 Feb 8 '17 at 23:20
  • $\begingroup$ After a little thought one sees that for definition 0 to satisfy def. 1, a sufficient condition is that "if a map $m'$ equalizes $c_1, c_2$ then it factorizes $f=m'e'$" and conversely, that if $f=m' e'$ with $m$ monic, then $m$ equalizes $c_1, c_2$"... $\endgroup$ – Noix07 Feb 8 '17 at 23:28

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