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$\require{AMScd}$In Tensor Categories by Etingof et al, a subobject $B$ of $A$ is not defined via isomorphism classes of monomorphisms. Rather, it is stated that a monomorphism $B \hookrightarrow A$ makes $B$ a subobject.

My first problem becomes now apparent:

  • They define a simple object $X$ to be an object that has only $0$ and itself as subobject. This seems already problematic to me, because isomorphisms break this definition. That is, if two objects $X$ and $Y$ are isomorphic, then they will not be simple using this definition. Defining simple object (or rather subobject) via isomorphism classes would get rid of this.

Now we want to prove Schur's Lemma (p 5), which is stated as

Let $X, Y$ be simple objects. Then any nonzero morphism $f\in\mathrm{Hom}[X,Y]$ is an isomorphism.

So far, my proof is as follows:

We have the canonical decomposition (unlabeled arrows are identities)

$$\begin{CD} K @>\ker f>>X @>\mathrm{coker }\ker f>> \mathrm{Im}(f) @>\ker\mathrm{coker } f>> Y @>\mathrm{coker }f>> C\\ @. @VVV @. @AAA\\ @. X @>f>>Y @>>>Y \end{CD}$$ Kernels are monic, so $K$ is either 0 or $X$, since $X$ is simple. It can't be $X$, since $f$ is supposed to be nonzero. But this means that $f$ is monic, so $\mathrm{Im}(f)=X$, $f= \ker \mathrm{coker } f$, and $\mathrm{coker }\ker f = \mathrm{id}_X$.

Now we get the problem again: It also implies that $X$ is a subobject of $Y$, which contradicts either the definition or the supposition.

My second problem is

  • What to do now? Can the proof be salvaged yet?

So, in summary:

  1. Is the definition sound?

  2. If the answer to 1. is "Yes", then how to complete the proof? (it is clear that a definition using isomorphism classes would make the proof a lot easier)

Jo Mo
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  • So $X$ is isomorphic to $Y$ – user12580 May 05 '17 at 15:30
  • Why should that follow immediately? – Jo Mo May 05 '17 at 15:33
  • Since X is a nonzero subobject of Y and Y is simple – user12580 May 05 '17 at 15:35
  • This book is pretty good, thank you for introducing it to me. – user12580 May 05 '17 at 15:36
  • No, in our definition, this would actually imply that $X$ is equal to $Y$. Not isomorphic, but equal. That is a huge difference. Or am I mistaken? – Jo Mo May 05 '17 at 15:40
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    I think the definition should be understood as "isomorphism classes" rather than "monomorphism". Or the definition of simple object should include a "up to isomorphism" somewhere. – Arnaud D. May 05 '17 at 15:54
  • A subobject of $A$ is a monomorphism $B\to A$ or an isomorphism class of monos into $A$, but it should definitely not be: an object $B$, such that there exists a mono $B\to A$. After all, "subobject" should be a generalization of e.g. subgroups, but in this fashion you would be treating isomorphic (as groups) subgroups as the same, which you shouldn't. That's also not what is stated in the book. – Stefan Perko May 05 '17 at 16:05
  • The difference between subobjects as monos and as isomorphism classes of monos is definitely not essential (replacing isomorphy with equality or vice versa). – Stefan Perko May 05 '17 at 16:12
  • @StefanPerko: Hm, I certainly did not write that I consider a subobject to be an object s.t. $\exists$ a mono. In fact, I pretty much wrote down Definition 1.3.5 from the book: "An object together with a monomorphism" (I feel that definition does make it seem like a property of an object). I don't understand your last remark, why is the difference negligible? Thanks – Jo Mo May 06 '17 at 05:20
  • Now I think the problem of me not being able to complete the proof is that the canonical decomposition is unique up to unique isomorphism (i.e. universal property of image/coimage). – Jo Mo May 06 '17 at 08:17
  • @JoBe That is how I understood your definition. If it is just me, then forget I said anything. - I can't think of any good reason why the difference is not negligible (we are replacing a thin category with a skeleton thereof; these categories are equivalent, so I consider them "the same") – Stefan Perko May 06 '17 at 12:39

1 Answers1

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Since in category theory everything is defined up to isomorphism I think that the authors of the book meant that

A simple object $X$ is an object whose subobjects, up to isomorphism, are just $0$ and $X$ itself.

About the second question I do not see any problem: you have proven so far that $X$, or to be exact $f \colon X \to Y$ is a subobject of $Y$, now since $Y$ is simple either $X$ is (isomorphic to) $0$ or it is (isomorphic to) $Y$. Since by hypothesis $f$ is not the null morphism clearly $X$ is not $0$ hence the thesis easily follows.

Giorgio Mossa
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