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I'm working on this exercise:

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(a) Here's how I understand (a) - is that right?

Suppose $im(m)\ne im(m')$. WLOG suppose there is $y\in im(m')$ but $y\notin im(m)$. Say $m'(x)=y$. If $\alpha:X'\to X$ is any map that makes the triangle commute, then we must have $m(\alpha(x))=y$, which contradicts the fact that $y\notin im(m)$. So there are no morphisms $(X,m')\to (X,m)$ at all, let alone bijective ones. This part doesn't seem to use the fact that $m,m'$ are monic.

Conversely, suppose $im(m)=im(m')$. Since $m,m'$ are monic, they are injective, so for every $y\in im(m)$ there is a unique $x\in X$ and $x'\in X'$ that are mapped to $y$. This enables one to establish a bijection between $X$ and $X'$ in a way that the triangle commutes.

So $m$ is isomorphic to $m'$ iff $im(m)=im(m')$.

The "canonical one-to-one correspondence" is, I believe the following. Given an equivalence class of an arrow $m: X\to A$, map it to $im(m)$. This is well-defined by the above result. This map is surjective: the preimage of $S\subset A$ is the map $S\to A$ given by $x\mapsto x$. Injectivity is clear too: if two equivalence classes are mapped to the same set, then consider one representative from each class; their images are the same, so they lie in the same class.

(b) Common sense suggests that the answer should be "subgroups, subrings, vector subspaces". In all these categories, monics are the same as injective homomorphisms. This being so, doesn't the same argument as above apply (according to which subobjects are again subsets)? Where should the requirement on closure under operations should come from?

(c) As far as I understand, the discussion here implies that in $\mathbf {Top}$ monics are the same as injective continuous maps. So the same question arises as in (b).

William Elliot
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user557
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1 Answers1

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a) looks good to me

Regarding b) and c) you want to have at least a faithful functor to $\mathsf{Set}$ to use the result of a). However note, that injective continuous functions give the wrong notion of subobject in $\mathsf{Top}$, since they don’t require the subobject to have the subspace topology. A better notion uses (isomorphism classes of) regular monomorphisms, i.e. monomorphisms coming from equalizers. In the cases listed in b) (edit except rings) these notions coincide, so you indeed have subgroups and subvectorspaces. I hoped this would set the question, where the closure properties come from.

Jonas Linssen
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  • The notions of monomorphisms and regular monomorphisms don't coincide in the category of rings. – Arnaud D. Feb 19 '20 at 09:14
  • Are you sure? I mean thanks to adjunction business we have an explicit description of equalizers in rings as the subring of elements mapping to the same. You claim not every subring arises in this form? – Jonas Linssen Feb 19 '20 at 09:53
  • The injection $i\colon \mathbb{Z}\to \mathbb{Q}$ is an epimorphism of rings. If it was an equalizer of two maps $u,v\colon \Bbb Q\to R$, you would necessarily have $u=v$, and thus $i$ would have a section and thus be an isomorphism. More generally, and category that is not balanced will have monomorphisms that are not regular. – Arnaud D. Feb 19 '20 at 10:01
  • Mhh. Thanks for pointing this out. Seems like I was too enthusiastic about the niceness of algebraic categories :/ – Jonas Linssen Feb 19 '20 at 10:09