Negligible Temperley–Lieb morphisms form a tensor ideal

Let $f:m\to n$ and $g:n\to m$ be basis diagrams. The closed matching graphs defining ${\mathrm{Tr}}_m(g\circ f)$ and ${\mathrm{Tr}}_n(f\circ g)$ are the same finite graph: the two stacked layers are read cyclically from opposite interfaces. They therefore have the same number of closed components and the same coefficient ${\delta }^c$. Bilinearity gives ${\mathrm{Tr}}_m(g\circ f)={\mathrm{Tr}}_n(f\circ g)$ for all such rectangular morphisms.

Suppose $f:m\to n$ is negligible, take arbitrary $a:r\to m$ and $b:n\to s$, and test $b\circ f\circ a:r\to s$ against any $h:s\to r$. The composite $a\circ h\circ b$ has type $n\to m$, so [F3] gives ${\mathrm{Tr}}_m(a\circ h\circ b\circ f)=0$. By step 1.1, this equals ${\mathrm{Tr}}_r(h\circ b\circ f\circ a)$. Thus $b\circ f\circ a$ is negligible, proving closure under composition on both sides.

Let $u:r\to s$ be arbitrary and let $h:n+s\to m+r$ test $f\otimes u:m+r\to n+s$. Form $g:n\to m$ by composing $1_n\otimes u$ with $h$ and then closing the last $r$ auxiliary input/output points; each resulting closed component contributes its factor $\delta$. The full closed matching graph for $h\circ (f\otimes u)$ has exactly the same remaining components and the same removed components as the graph obtained by first making this partial closure and then forming $g\circ f$. Therefore ${\mathrm{Tr}}_{m+r}(h\circ (f\otimes u))={\mathrm{Tr}}_m(g\circ f)=0,$ where the last equality is [F3]. This proves $f\otimes u$ negligible.

For $u\otimes f:r+m\to s+n$, let a test morphism be $h:s+n\to r+m$. Compose $u\otimes 1_n$ with $h$ and close the first $r$ auxiliary input/output points to obtain $g:n\to m$. The same component count gives ${\mathrm{Tr}}_{r+m}(h\circ (u\otimes f))={\mathrm{Tr}}_m(g\circ f)=0$. Thus $u\otimes f$ is negligible as well, proving tensor stability on both sides.