
Lemma 10.11.5. The homomorphisms

$(M \otimes _ R N)\otimes _ R P \to M \otimes _ R N \otimes _ R P \to M \otimes _ R (N \otimes _ R P)$

such that $f((x \otimes y)\otimes z) = x \otimes y \otimes z$ and $g(x \otimes y \otimes z) = x \otimes (y \otimes z)$, $x\in M, y\in N, z\in P$ are well-defined and are isomorphisms.

Proof. We shall prove $f$ is well-defined and is an isomorphism, and this proof carries analogously to $g$. Fix any $z\in P$, then the mapping $(x, y)\mapsto x \otimes y \otimes z$, $x\in M, y\in N$, is $R$-bilinear in $x$ and $y$, and hence induces homomorphism $f_ z : M \otimes N \to M \otimes N \otimes P$ which sends $f_ z(x \otimes y) = x \otimes y \otimes z$. Then consider $(M \otimes N)\times P \to M \otimes N \otimes P$ given by $(w, z)\mapsto f_ z(w)$. The map is $R$-bilinear and thus induces $f : (M \otimes _ R N)\otimes _ R P \to M \otimes _ R N \otimes _ R P$ and $f((x \otimes y)\otimes z) = x \otimes y \otimes z$. To construct the inverse, we note that the map $\pi : M \times N \times P \to (M \otimes N)\otimes P$ is $R$-trilinear. Therefore, it induces an $R$-linear map $h : M \otimes N \otimes P \to (M \otimes N)\otimes P$ which agrees with the universal property. Here we see that $h(x \otimes y \otimes z) = (x \otimes y)\otimes z$. From the explicit expression of $f$ and $h$, $f\circ h$ and $h\circ f$ are identity maps of $M \otimes N \otimes P$ and $(M \otimes N)\otimes P$ respectively, hence $f$ is our desired isomorphism. $\square$

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