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?
Edit:
Here are two examples for non-semisimple finite tensor categories with simple projective objects:
- The symplectic fermions.
This category is even factorizable, i.e. ribbon with a certain non-degeneracy condition on the braiding. - The category of representations over the restricted quantum group $\overline{U}_q(sl_2)$.
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.
Best Answer
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.