# Supremum of a family of subobjects in an abelian category

So this question follows on from an earlier one here. Suppose we are in a cocomplete abelian category. Say we have a (small) family of subobjects of some object $A$, which we will write as $\{ \mu_{i} : A_{i} \hookrightarrow A \}_{i \in I}$. Define the categorical sum as the image of the canonical morphism $$\phi: \bigoplus_{i \in I} A_{i} \longrightarrow A.$$ Order the subobjects by inclusion: That is $A_{i} \leq A_{j}$ if there is a monomorphism $u: A_{i} \rightarrow A_{j}$ compatible with the inclusions $\mu_{i}$ and $\mu_{j}$). Then it is not hard to find that $\text{im}(\phi)$ is a subobject of $A$ and provides an upper bound for this family under the partial ordering. However, it seems to be assumed in the literature that this is in fact a supremum. Is anyone able to help me show this? In other words, if $B$ is some other subobject of $A$ acting as an upper bound, I need to find a morphism, $$\rho: \text{im} (\phi) \longrightarrow B$$ compatible with the subobject structure. I feel like this should follow from universal properties of the image and direct sum, but I haven't been able to get it. I also tried to use epi-mono factorization properties. Is it actually true if the family isn't directed? Any help is appreciated.

Since $B$ is an upper bound for each $A_i$, they all factor through $B$: there are maps $\nu_i:A_i\to B$ such that the composition $A_i\to B\to A$ is the inclusion $A_i\to A$. These maps $\nu_i$ combine to give a map $\nu:\bigoplus A_i\to B$ whose composition with $B\to A$ is $\phi$. Thus the image of $\phi$ factors through $B$.

• Thank you for the reply, this does help, although I am afraid I am still not seeing it. I following everything you say right up until the very last sentence. Why does this cause the image to factor through $B$? The image is the kernel of the cokernel, as least by my definition (is this the right one to be using here?). Are we using some property of kernels and cokernels here? What exactly is the universal property you're using to claim that it facts through $B$? I apologize if I am just being stupid here. – Luke Feb 12 '18 at 5:05
• This takes a fair amount of work to prove directly from the kernel of cokernel definition. If you know that the image is the unique epi-mono factorization, though, you can just take the image of $\nu$ and see that its composition with $B\to A$ gives another epi-mono factorization of $\phi$. – Eric Wofsey Feb 12 '18 at 5:09
• I see, that makes sense then. I suppose proving it directly would be more or less reproving the fact that the epi-mono factorization is unique anyway. – Luke Feb 12 '18 at 5:11
• Or, if you're like me and like being lazy and citing big theorems, you can just use the Freyd-Mitchell embedding theorem to assume without loss of generality you are in a category of modules, in which case the statement in question is obvious. (To be precise, the statement in question is that if $\nu:X\to Y$ and $i:Y\to Z$ with $i$ monic, then the image of $i\nu$ factors through $i$.) – Eric Wofsey Feb 12 '18 at 5:12
• Actually this is quite clever. I could appeal to a diagram chase then. As a statement about elements of a set, yes, it is quite obvious. Thanks! – Luke Feb 12 '18 at 5:25

It is true !

Assume $\iota : B\hookrightarrow A$ is a subobject and for each $i\in I$ we have an inclusion $\delta_i : A_i\hookrightarrow B$ such that the obvious diagram commutes ($\iota\circ \delta_i = \mu_i$)

Then we also get an induced morphism $\psi: \displaystyle\bigoplus_{i\in I} A_i \to B$. But of course, since the morphisms were coherent in the beginning, the uniqueness part of the definition of coproduct makes it clear that $\iota\circ \psi = \phi$. Indeed, for each $i\in I$, $\iota\circ \psi\circ m_i = \iota\circ \delta_i= \mu_i = \phi\circ m_i$, where $m_i : A_i \to \displaystyle\bigoplus_{j\in I} A_j$ is the inclusion.

$\iota \circ \psi = \phi$, with $\iota$ monic. So if we write $\psi$ in its epi-mono-factorisation, we get an epi-mono factorisation for $\phi$. These are essetially unique and so $\text{im}(\phi) \hookrightarrow A$ factors through this, in particular it factors through $B$.