$\require{AMScd}$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)

  • $\begingroup$ So $X$ is isomorphic to $Y$ $\endgroup$ – user12580 May 5 '17 at 15:30
  • $\begingroup$ Why should that follow immediately? $\endgroup$ – Jo Be May 5 '17 at 15:33
  • $\begingroup$ Since X is a nonzero subobject of Y and Y is simple $\endgroup$ – user12580 May 5 '17 at 15:35
  • $\begingroup$ This book is pretty good, thank you for introducing it to me. $\endgroup$ – user12580 May 5 '17 at 15:36
  • 1
    $\begingroup$ 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. $\endgroup$ – Arnaud D. May 5 '17 at 15:54

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.


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