Lemma 10.12.5. The homomorphisms
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.
Lemma 10.12.5. The homomorphisms
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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