# Subobject of an element in a category

I am trying to understand what is a subobject of an element $A$ in a category $\mathcal{C}$. I am reading English translation some aspects of homological algebra of "Sur quelques points d'algèbre homologique" by Alexander Grothendieck.

Discussion goes as follows :

Given monomorphisms $u:B\rightarrow A$ and $u':B'\rightarrow A$, we say $u'$ majorizes $u$ and write $u\leq u'$ if $u$ factors through $u'$ i.e., there exists $v:B\rightarrow B'$ such that $B\xrightarrow{v}B'\xrightarrow{u'}A$ is the map $u:B\rightarrow A$.

We say $u,u'$ are equivalent if $u\leq u',u'\leq u$ and $v:B\rightarrow B'$ and $v':B'\rightarrow B$ are inverses of each other.

Then he says

choose one monomorphism in each class of equivalent monomorphisms, the selected monomorphims will be called the subobjects of $A$.

This equivalenece relation partitions the collection of all monomorphisms. Suppose $u'$ majorizes $u$ but $u$ does not majorize $u'$ then $u$ will not be in the equivalence class of $u'$ but $u$ will be in another equivalence class which it represents and this will be then called a subobject too. This seems absurd.

I am surely missing something simple. Any suggestions are welcome.

• The equivalence relation in question, insofar as there is one of any significance, is "$u\leq u'\wedge u'\leq u$". $x\textrm{ majorizes } y$ is not intended to be an equivalence relation. – Malice Vidrine Aug 3 '17 at 17:32
• I am not a native speaker of english so, I do not fully understand your comment @MaliceVidrine – user312648 Aug 3 '17 at 17:39
• Perhaps I've misunderstood your question, but it sounds like you've been given a definition of equivalence as "$u\leq u'\wedge u'\leq u$" and are wondering why, when you suppose $u\leq u'$ and not $u'\leq u$,they should be in different equivalence classes. – Malice Vidrine Aug 3 '17 at 17:48
• In either case, what's being described here is the quotienting of a preorder into a partial order. This is important because given an object in an arbitrary category, the monomorphisms into it may be a proper class, but the quotient described here will very often be a set. And one can then meaningfully talk about a functor $Sub(-):\mathcal{C}\to \mathbf{Set}$. – Malice Vidrine Aug 3 '17 at 17:53
• @MaliceVidrine : This is what I am looking for. The reason behind considering the equivalence classes. Can you take your own time and make it an answer – user312648 Aug 3 '17 at 18:01

The construction here is to take the preorder of monomorphisms into some object and reduce it to a partial order (the elements of which we call the "subobjects"). The reason we want to do this is because the preorder of monomorphisms can be a proper class, but the partial order will often be a set. Think of how many monomorphisms there are into a given non-empty set $A$, compared to how many subsets there are of $A$.
The reason we want to do this is that when the subobjects of the objects of a finitely complete category $\mathcal{C}$ form a set, we can define a functor $Sub(-):\mathcal{C}^{op}\to \mathbf{Set}$. It is in terms of this functor that we can most easily start talking about the internal logic of a category, and is of key interest in topos theory. For example, $\mathcal{C}$ has a subobject classifier if and only if $Sub(-)$ is representable.
Think about the category of sets, a subobject of $A$ is a subset $B\subset A$ which can be seen as the equivalence class of an injective map $i_B:B\rightarrow A$, the fact that $i_C\leq i_B$ is equivalent to say that you have $f:C\rightarrow B$ such that $i_C=i_B\circ f$, $i_B\leq i_C$ and $i_C\leq i_B$ is equivalent to saying that $f$ is an isomorphism, you can have $C\subset B\subset A$ and $A,B,C$ are different. It is not absurd.