1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $Y$ ? $\endgroup$ – Paul Slevin Dec 17 '12 at 17:36
  • $\begingroup$ I have just realised that is true, because we are talking about subobjects and hence any factorisation must be a mono (and unique). $\endgroup$ – Paul Slevin Dec 17 '12 at 18:10
  • $\begingroup$ 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? $\endgroup$ – jennifer Aug 2 '17 at 0:51
  • $\begingroup$ 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. $\endgroup$ – jennifer Aug 2 '17 at 0:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.