The reduced Burau homology module

Each deck transformation induces an automorphism. By [F3] each $T_{t^k}$ is a homeomorphism, so by [F2] it induces $(T_{t^k}{)}_{\ast }:=H_1((T_{t^k}{)}_{\mathrm{\#}})$, an endomorphism of $M_{\mathrm{r}\mathrm{e}\mathrm{d}}$. Since $T_{t^k}\circ T_{t^{-k}}=\mathrm{id}=T_{t^{-k}}\circ T_{t^k}$ by [F3], functoriality in [F2] gives $(T_{t^k}{)}_{\ast }\circ (T_{t^{-k}}{)}_{\ast }=\mathrm{id}$ and $(T_{t^{-k}}{)}_{\ast }\circ (T_{t^k}{)}_{\ast }=\mathrm{id}$; hence each $(T_{t^k}{)}_{\ast }$ is an automorphism of $M_{\mathrm{r}\mathrm{e}\mathrm{d}}$, and $t_{\ast }:=(T_t{)}_{\ast }$ is a unit of the ring ${\mathrm{End}}_{\mathbb{Z}}(M_{\mathrm{r}\mathrm{e}\mathrm{d}})$ with inverse $(T_{t^{-1}}{)}_{\ast }$.

The ${\mathrm{\Lambda }}_1$-module structure. By [F1] applied directly to the unital ring ${\mathrm{End}}_{\mathbb{Z}}(M_{\mathrm{r}\mathrm{e}\mathrm{d}})$, where the inverse of $t_{\ast }$ was proved in step 1.1, the unit $t_{\ast }$ determines a unique unital ring homomorphism $\phi :{\mathrm{\Lambda }}_1⟶{\mathrm{End}}_{\mathbb{Z}}(M_{\mathrm{r}\mathrm{e}\mathrm{d}}),t⟼t_{\ast },$ and we let ${\mathrm{\Lambda }}_1$ act on $M_{\mathrm{r}\mathrm{e}\mathrm{d}}$ through $\phi$. This gives $M_{\mathrm{r}\mathrm{e}\mathrm{d}}$ the structure of a left ${\mathrm{\Lambda }}_1$-module, since ${\mathrm{\Lambda }}_1$ is commutative; it is well defined because $\phi$ is unique, so no further choice enters.

The action of $t^k$ is the induced automorphism. Since $\phi$ is a unital ring homomorphism, $\phi (t^k)=\phi (t{)}^k=(t_{\ast }{)}^k$ for every $k\in \mathbb{Z}$, and by functoriality of [F2] applied to the composition of $T_t$ with itself, $(t_{\ast }{)}^k=(T_{t^k}{)}_{\ast }$. Hence $t^k$ acts on $M_{\mathrm{r}\mathrm{e}\mathrm{d}}$ exactly by $(T_{t^k}{)}_{\ast }$, as asserted, and in particular the action is the deck action and not an additional datum. No choice principle is used.