# Smallest subobject in an abelian category containing a set of objects

Let $\mathcal A$ be an abelian category. Let $A$ be an object in $\mathcal A$ and let $(A_i)_{i\in I}$ be a set of subobjects of $A$. Then there is a subobject $\sum_{i \in I} A_i$ of $A$ which has all the $A_i$'s as a subobject, namely the image of the canonical morphism $\gamma \colon \coprod_{i \in I} A_i \to A$. Please refer to this question for more on this. However I am having trouble showing that this is the smallest such subobject.

What does "smallest subobject" mean in this sense? Does it mean given any subobject $X \longrightarrow A$ such that every $A_i$ is a subobject of $X$ then $$\operatorname{im} \gamma \longrightarrow A \le X \longrightarrow A$$ OR does it mean that given any subobject $X \longrightarrow A$ such that every $A_i$ is a subobject of $X$ and also we have a commutative diagram of subobject arrows $$\begin{array}{ccc} A_i & \rightarrow & A \\ & \searrow & \uparrow \\ & & X \end{array}$$ then $$\operatorname{im} \gamma \longrightarrow A \le X \longrightarrow A ?$$

In the latter case I can see it because we can just use the universal property of the coproduct and the image to draw an arrow $\operatorname{im}\gamma \to X$ which works. However I cannot see this for the former case. If "smallest subobject" means the latter case why is it not completely specified in the language; is it a convention?

Let us say that $i:A\to X$ factors through $B$ if $j:B\to X$ is a subobject and there exists $k:A\to B$ such that $i=j\circ k$.
You are looking for a subobject $X\to A$ which has the following property: Whenever $A_i\to A$ factors through $Y$ for all $i$, then $X$ factors through $Y$ as well. In other words, whenever you have a subobject $Y$ "containing" all $A_i$, then $Y$ "contains" $X$. Now this property is satisfied if you take for $X$ the image of $\coprod_i A_i\to A$ simply by the universal property of the image.
• I misunderstood "contains $A_i$" to mean "have $A_i$ as a subobject." Thanks for clarifying. Is it necessarily true that $A_i$ will be a subobject of some $Y$ if it factors through $Y$ ? – Paul Slevin Dec 17 '12 at 17:36
• I'm doing this exercise now, and I can't figure out how to apply the universal property of the image. More specifically, if $A_i\rightarrow A$ factors through $Y$ for all $i$, then there's a map $\coprod A_i\rightarrow Y$ induced from the U.P. for coproduct. However, to apply U.P. for image, I need for $\coprod A_i\rightarrow A$ to have equal factorizations $\coprod A_i\rightarrow I\rightarrow A$ and $\coprod A_i\rightarrow Y\rightarrow A$. How do I show this? – jennifer Aug 2 '17 at 0:51
• I'm getting $A_i\rightarrow\coprod A_i\rightarrow I\rightarrow A$ and $A_i\rightarrow \coprod A_i\rightarrow Y\rightarrow A$ are the same, but $A_i\rightarrow \coprod A_i$ isn't (necessarily) epi. So I can't conclude what I want immediately. – jennifer Aug 2 '17 at 0:54