Associator naturality, pentagon, unit triangle and symmetry hexagons on elementary tensors

Naturality of $\alpha$ and $\sigma$: for linear maps $f:L\to L^{\mathrm{\prime }}$, $g:M\to M^{\mathrm{\prime }}$, $h:N\to N^{\mathrm{\prime }}$ and an elementary tensor $(l\otimes m)\otimes n$ one has ${\alpha }_{L^{\mathrm{\prime }},M^{\mathrm{\prime }},N^{\mathrm{\prime }}}(((f\otimes g)\otimes h)((l\otimes m)\otimes n))={\alpha }_{L^{\mathrm{\prime }},M^{\mathrm{\prime }},N^{\mathrm{\prime }}}((f(l)\otimes g(m))\otimes h(n))=f(l)\otimes (g(m)\otimes h(n))$ and $(f\otimes (g\otimes h))({\alpha }_{L,M,N}((l\otimes m)\otimes n))=(f\otimes (g\otimes h))(l\otimes (m\otimes n))=f(l)\otimes (g(m)\otimes h(n))$, by [F2] and [F4]; likewise ${\sigma }_{M^{\mathrm{\prime }},N^{\mathrm{\prime }}}((g\otimes h)(m\otimes n))=h(n)\otimes g(m)=((h\otimes g)\circ {\sigma }_{M,N})(m\otimes n)$. Both sides of each square are $k$-linear and the elementary tensors span by [F5], so the squares commute on their whole domains.

Naturality of the unit isomorphisms: for linear $f:N\to N^{\mathrm{\prime }}$ and $r\otimes n\in k\otimes N$ one has ${\lambda }_{N^{\mathrm{\prime }}}(({\mathrm{i}\mathrm{d}}_k\otimes f)(r\otimes n))={\lambda }_{N^{\mathrm{\prime }}}(r\otimes f(n))=rf(n)=f(rn)=f({\lambda }_N(r\otimes n))$ by [F3] and [F4], and for linear $g:M\to M^{\mathrm{\prime }}$ likewise ${\rho }_{M^{\mathrm{\prime }}}((g\otimes {\mathrm{i}\mathrm{d}}_k)(m\otimes r))=g(m)r=g(mr)=g({\rho }_M(m\otimes r))$; the elementary tensors span, so both naturality squares commute.

Pentagon: on an elementary tensor $((l\otimes m)\otimes n)\otimes x$ of $((L\otimes M)\otimes N)\otimes X$ the left composite sends it by [F2] to $({\alpha }_{L,M,N}\otimes {\mathrm{i}\mathrm{d}}_X)(((l\otimes m)\otimes n)\otimes x)=(l\otimes (m\otimes n))\otimes x$, then to $l\otimes ((m\otimes n)\otimes x)$, then to $l\otimes (m\otimes (n\otimes x))$; the right composite sends it to $(l\otimes m)\otimes (n\otimes x)$ and then to $l\otimes (m\otimes (n\otimes x))$. Both sides are $k$-linear maps whose domain is spanned by such elementary tensors [F5], so the two composites agree everywhere.

Unit triangle: for an elementary tensor $(m\otimes c)\otimes n$ of $(M\otimes k)\otimes N$ the left side gives $({\mathrm{i}\mathrm{d}}_M\otimes {\lambda }_N)(m\otimes (c\otimes n))=m\otimes (cn)$ by [F2] and [F3], while the right side gives $({\rho }_M\otimes {\mathrm{i}\mathrm{d}}_N)((m\otimes c)\otimes n)=(mc)\otimes n$; these are equal because $(mc)\otimes n=m\otimes (cn)$ by the balancing relations of [F5]. Both sides are linear on the span of the elementary tensors, so the identity holds.

First symmetry hexagon: on an elementary tensor $(l\otimes m)\otimes n$ of $(L\otimes M)\otimes N$ the left composite gives successively $n\otimes (l\otimes m)$ (symmetry ${\sigma }_{L\otimes M,N}$), $(n\otimes l)\otimes m$ (inverse associator), $(l\otimes n)\otimes m$ (symmetry in the first factor), $l\otimes (n\otimes m)$ (associator), while the right composite gives $l\otimes (m\otimes n)$ and then $l\otimes (n\otimes m)$ by the symmetry in the second factor; the two agree on the spanning elementary tensors, hence everywhere.

Second symmetry hexagon: on an elementary tensor $l\otimes (m\otimes n)$ of $L\otimes (M\otimes N)$ the left composite gives $(l\otimes m)\otimes n$, then $n\otimes (l\otimes m)$, then $(n\otimes l)\otimes m$, while the right composite gives $l\otimes (n\otimes m)$, then $(l\otimes n)\otimes m$, then $(n\otimes l)\otimes m$; agreement on the spanning elementary tensors gives the identity everywhere.

Steps 1.1–1.6 verify all five identities on elementary tensors, and each identity is between $k$-linear maps whose domains are the iterated tensor products of [F1] spanned by elementary tensors [F5]; a linear map is determined by its values on a spanning set, so each identity holds on its whole domain, and no general monoidal coherence theorem was invoked.