A Jones–Wenzl annihilator exists exactly when the trivial module is projective

A basis diagram has $n$ propagating strands exactly when it is the identity. If either factor in a product has fewer than $n$ propagating strands, every exterior through path in the product must pass through a distinct propagating path in that factor; hence the product also has fewer than $n$. Thus the span of nonidentity diagrams is a two-sided ideal and $\epsilon$ is an algebra homomorphism. Its kernel is spanned by nonempty generator words by [F1].

If $K_{\epsilon }$ is projective, the surjection $\epsilon :A\to K_{\epsilon }$ splits as a left $A$-module map $s$. Put $e=s(1)$. Then $\epsilon (e)=1$ and, for every $a\in A$, $ae=s(\epsilon (a))=\epsilon (a)e$. Taking $a=e$ gives $e^2=e$, and taking $a=d_i$ gives $d_{i}e=0$.

Conversely suppose $e\mathrm{\ne }0$, $e^2=e$, and $d_{i}e=0$ for every generator. Every nonempty generator word kills $e$ by its rightmost letter, so [F1] gives $ae=\epsilon (a)e$ for all $a\in A$. If $\epsilon (e)=0$, then $e\in \ker \epsilon$ is a linear combination of nonempty generator words, whence $e^2=0$, contradicting $e^2=e\mathrm{\ne }0$. Thus $\epsilon (e)=1$, and $s:K_{\epsilon }\to A$, $s(\lambda )=\lambda e$, is an $A$-linear section of $\epsilon$. Hence $K_{\epsilon }$ is a direct summand of the free left regular module and is projective.

Steps 2.1 and 2.2 prove both implications and the normalization $\epsilon (e)=1$. No uniqueness assertion is needed for this projectivity equivalence.