Lemma 8.10.5 in EGNO's Tensor Categories basically states

Let $\mathcal{C}$ be a tensor category over an algebraically closed field $\mathbb{k}$ with braiding $c$. For any nonzero simple object $X$ the composition \begin{align} t(X) := \operatorname{ev}_X \circ c_{X, X^\vee} \circ \operatorname{coev}_X \in \operatorname{End}_{\mathcal{C}}(\mathbf{1}) \end{align} is nonzero.

I feel very conflicted. On the one hand, the one line proof given in the book seems plausible:

Since $X$ is simple, the corresponding composition \begin{align} \operatorname{End}(\mathbf{1}) \to \operatorname{Hom}(\mathbf{1}, X\otimes X^\vee) \to \operatorname{End}(\mathbf{1}) \end{align} consists of nonzero maps between 1-dimensional spaces, and is thus non-zero.

On the other hand, suppose that the lemma holds and that $X$ is projective. Then $P = X \otimes X^\vee$ is projective. Set $f = t(X)^{-1} \operatorname{coev}_X$ and $g = \operatorname{ev}_X \circ c_{X, X^\vee} $. But then \begin{align} \mathbf{1} \xrightarrow{f} P \xrightarrow{g} \mathbf{1} = \operatorname{id}_{\mathbf{1}} \ , \end{align} so that $\mathbf{1}$, being a direct summand in a projective, is projective. But then $\mathcal{C}$ is semisimple. A contradiction to the existence of non-semisimple finite tensor categories with simple projective objects.

Note that in fact the general heuristic in this last part implies that in a non-semisimple (finite) tensor category there exists no nonzero endomorphism of the tensor unit factoring through a projective object. For this heuristic, see also the proof of Theorem 6.6.1 in the book.

So, where is the mistake?


Here are two examples for non-semisimple finite tensor categories with simple projective objects:

Edit 2: The mistake is in the proof in the book. Namely, as I prove, the map $\operatorname{Hom}(\mathbf{1}, X \otimes X^\vee) \to \operatorname{End}(\mathbf{1})$ is zero if $X$ is projective.

  • 1
    $\begingroup$ I don't know any examples of non-semisimple finite tensor categories with simple projective objects off the top of my head, but I think what you've written down is a proof by contradiction that no such thing can have a braiding, yes? $\endgroup$ Sep 15, 2020 at 8:07
  • $\begingroup$ @QiaochuYuan: I added two examples. Unfortunately, one of the categories I list definitely is braided. So I have absolutely no idea what's going on :D $\endgroup$
    – Jo Mo
    Sep 15, 2020 at 8:17
  • $\begingroup$ @QiaochuYuan: Actually, there also exists a version of the second example where the coproduct is modified so that the category also becomes factorizable. Of course this is not twist equivalent to my example. $\endgroup$
    – Jo Mo
    Sep 16, 2020 at 4:48
  • 1
    $\begingroup$ Please mail Etingof, who keeps a list of errata on his website. $\endgroup$ Sep 16, 2020 at 9:32
  • $\begingroup$ @darijgrinberg I just did, thanks $\endgroup$
    – Jo Mo
    Sep 16, 2020 at 9:54

1 Answer 1


The deceptively simple proof in the book indeed managed to deceive us.

How? It assumes that the linear map \begin{align} \operatorname{Hom}(\mathbf{1}, X^\vee \otimes X) &\to \operatorname{End}(\mathbf{1}) \newline f &\mapsto \operatorname{ev}_X \circ f \end{align} is non-zero, which is not true according to my proof above.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .