Let $A$ be an Algebra, $S$ a subalgebra ans $W$ a subspace s.t $SW \subset W$, $WS \subset W$ and $S,W$ generate $A$. Show that for any $A-$module $M$ the follow sequence is exact.
$$A \otimes_S W \otimes_S M\stackrel{f_1} {\longrightarrow} A\otimes_S M \stackrel{f_0} \longrightarrow M \longrightarrow 0, $$
where $f_1: A \otimes_S W \otimes_S M \to A\otimes_S M$ is given by $a\otimes w \otimes m \mapsto a w \otimes m - a \otimes wm$ and $f_0: A\otimes_S M \to M $ is given by $a \otimes m \mapsto am.$
Well, I have to show that $ Im f_1 = \ker $$f_0$ and $f_0$ is surjective. Let $m \in M$, we have $m = f_0(1 \otimes m)$, so $f_0$ is surjective. Actually, I guess $f_0$ is bijective, where $f_0^{-1}$ is given by $m \mapsto 1\otimes m$. So, if it is correct then $\ker f_0 = \{0\}$, but it could be $Im f_1??$
I Guess zero in $A \otimes_SM$ is given by the space generated by $a\otimes w \otimes m \mapsto a w \otimes m - a \otimes wm$.
One way to prove this is as follows. This has the advantage of not having to work with elements, and the exactness comes for free.
Let $T$ be the tensor algebra of $W$ over $S$, so $T=T_0\oplus T_1\oplus T_2\oplus\cdots$ is a graded algebra, where $T_0=S$ and $T_{n+1}=T_n\otimes_SW$. The multiplication comes from concatenation of tensors. A basic example of this construction is the path algebra of a quiver: take a field $K$ and a quiver $Q$, and set $S=KQ_0\cong K^n$, having one idempotent for each vertex, and $W=KQ_1$, having $K$-basis given by the arrows and the natural $S$-bimodule structure.
We set $T_+=T_1\oplus T_2\oplus\cdots$ for the graded radical, so that $T/T_+\cong S$ and $T_+\cong T\otimes_SW$ as a $T$-$S$-bimodule. Then $T=S\oplus T_+$ and we have the split exact sequence of $S$-bimodules $$ 0 \to T_+ \xrightarrow{\iota} T \xrightarrow{\pi} S \to 0. $$
Given a $T$-module $M$, we can tensor the above sequence to obtain a split exact sequence of $S$-modules $$ 0 \to T_+\otimes_SM \xrightarrow{\iota} T\otimes_SM \xrightarrow{\pi} M \to 0. $$ We now observe that the $S$-module map $\theta\colon T\otimes_SM\to T\otimes_SM$, given on $T_{n+1}\otimes_SM\cong T_n\otimes_SW\otimes_SM$ by $$ \theta(t_n\otimes w)\otimes m := t_n\otimes(wm), $$ is locally nilpotent, so $\delta:=1-\theta$ is invertible. Setting $g_0:=\pi\delta^{-1}$ and $g_1:=\delta\iota$ yields an exact sequence $$ 0 \to T_+\otimes_SM \xrightarrow{g_1} T\otimes_SM \xrightarrow{g_0} M \to 0 $$ which is in fact now an exact sequence of left $T$-modules.
If $T=KQ$ is the path algebra of a quiver, then this is a projective resolution of $M$, often called the standard projective resolution of $M$.
Finally, your algebra $A$ is a quotient of $T$ by some ideal $I$ (intersecting $T_0\oplus T_1$ trivially). So if $M$ is an $A$-modules, then we apply $A\otimes_T-$ to the above sequence to obtain an exact sequence $$ A\otimes_TT_+\otimes_SM \xrightarrow{f_1} A\otimes_SM \xrightarrow{f_1} M \to 0. $$ Now use that $T_+\cong T\otimes_SW$ to get $$ A\otimes_TT_+\otimes_SM \cong A\otimes_SW\otimes_SM. $$