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A concrete category is a category $C$ endowed with a faithful functor $U:C\rightarrow Set$. And if $a$ is an object in $C$, then a subobject of $a$ is an isomorphism class of monomorphisms with codomain $a$.

I’m wondering if subobjects can be related to subsets using $U$. My question is, given a subobject of $a$, does there always exist a monomorphism $f$ in the subobject such that $U(f)$ is an inclusion map from a subset of $U(a)$ to $U(a)$?

If not, does anyone know of a counterexample?

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  • $\begingroup$ I don't understand the question; what does "in the subobject" mean? The question that it seems most likely you're asking is: does $U$ preserve monomorphisms? The answer is not always, but often yes in practice, and under mild hypotheses (e.g. $U$ preserving finite limits). $\endgroup$ – Qiaochu Yuan Mar 12 at 1:45
  • $\begingroup$ @QiaochuYuan Given an object $a$ in $c$, consider the class $X$ of all monomorphisms whose codomain is $a$, and let us define an equivalence relation $\sim$ on $X$ as follows: $f\sim g$ if there exists an isomorphism $h$ In $C$ such that $f=gh$. And let us take equivalence classes of elements of $X$ under $\sim$. These equivalence classes are called subobjects of $a$. I hope that makes it clear what it means for a monomorphism to be an element of a subobject. $\endgroup$ – Keshav Srinivasan Mar 12 at 3:22
  • $\begingroup$ @QiaochuYuan See here and here for more details. And no, I’m not asking whether $U$ maps monomorphisms to monomorphisms. To use the terminology of my previous comment, I’m asking, given a monomorphism $f\in X$, does there exist a monomorphism $g\in X$ such that $f\sim g$ and $U(g)$ is an inclusion map from a subset of $U(a)$ to $U(a)$? $\endgroup$ – Keshav Srinivasan Mar 12 at 3:30
  • $\begingroup$ You mean a monomorphism that represents a subobject? "I think "in the subobject" is extremely confusing terminology for this. Most people would just identify subobjects and monomorphisms by abuse of notation. Also, as in Kevin Carlson's answer, this is an extremely unnatural question to ask from the categorical point of view, because it fails to be invariant under equivalence of categories. See ncatlab.org/nlab/show/principle+of+equivalence for more. $\endgroup$ – Qiaochu Yuan Mar 12 at 8:35
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Consider the full subcategory $C$ of Set on exactly two objects: the natural numbers and the singleton set $b$ whose element is my coffee mug (if you prefer, let $b=\{\pi\}$.) $C$ comes equipped with the obvious forgetful functor $U$ which is just the inclusion. There are infinitely many distinct ways in which $b$ is a subobject of $\mathbb N$ in $C$, but there are no proper subsets of $\mathbb N$ in the image of $U$.

For a more general answer, observe that the existence of a faithful functor into Set is invariant under equivalence of categories. So $U$ can't possibly impose any on-the-nose conditions like yours.

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