N=\Delta(1)(M\ot N),$$ the $k$-submodule of $M\ot N$ generated by elements of the form $1_{(1)}\ot 1_{(2)}$. $M\ot_t N$ is a left $H$-module, with left diagonal action $h\cdot(m\ot n)=h_{(1)}m\ot h_{(2)}n$. It follows from [(\[eq:1.1.1\])]{} that the tensor product $\ot_t$ is associative. Observe that $$M\ot_t N\ot_t P=\Delta^2(1)(M\ot N\ot P).$$ $H_t\in {}_H{\mathcal{M}}$, with left $H$-action $h\leftact z=\varepsilon_t(hz)$. The induced $H_t$-bimodule structure is given by left and right multiplication by elements of $H_t$.\
For $M,N\in {}_H{\mathcal{M}}$, consider the projection $$\pi:\ M\ot N\to M\ot_t N,~~\pi(m\ot n)=1_{(1)}m\ot 1_{(2)}n.$$ Applying $\ol{\varepsilon}_s\ot H_t$ to [(\[eq:1.2b.1\])]{}, we find $$\ol{\varepsilon}_s(z\varepsilon_t(1_{(1)}))\ot 1_{(2)}=
1_{(1)}\ol{\varepsilon}_s(z)\ot 1_{(2)}=1_{(1)}\ot 1_{(2)}z,$$ hence $$\pi(mz\ot n)=\pi(\ol{\varepsilon}_s(z)m\ot n)=
1_{(1)}\ol{\varepsilon}_s(z)m\ot 1_{(2)}n
=1_{(1)}m\ot 1_{(2)}zn=\pi(m\ot zn).$$ So $\pi$ induces a map $\ol{\pi}:\ M\ot_{H_t} N\to M\ot_t N$, which is a left $H_t$-module isomorphism with inverse given by $$\ol{\pi}^{-1}(1_{(1)}m\ot 1_{(2)}n)=1_{(1)}m\ot_{H_t} 1_{(2)}n=m\ot_{H_t}n.$$

[\[pr:1.2c\]]{} Let $H$ be a weak bialgebra. Then we have a monoidal category $({}_H{\mathcal{M}},\ot_t,H_t,a,l,r)$. The associativity constraints are the natural ones. The left and right unit constraints $l_M:\ H_t\ot_t M\to M$ and $r_M:\ M\ot_t H_t\to M$ and their inverses are given by the formulas $$l_M(1_{(1)}\leftact z\ot 1_{(2)}m)=zm~~;~~l_M^{-1}(m)=\varepsilon_t(1_{(1)})\ot 1_{(2)}m;$$ $$r_M(1_{(1)}m\ot 1_{(2)}\leftact z)=\ol{\varepsilon}_s(z)m~~;~~
r_M^{-1}(m)=1_{(1)}m\ot 1_{(2)}.$$

This is a direct consequence of the observations made above. Let us check that $$\begin{aligned}
&&\hspace*{-2cm}
l_M^{-1}(l_M(1_{(1)}\leftact z\ot 1_{(2)}m))=l_M^{-1}(zm)=
\varepsilon_t(1_{(1)})\ot 1_{(2)}zm\\
&=&z\varepsilon_t(1_{(1)})\ot 