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I'm trying to verify for myself some isomorphisms of tensor products.

If $M$, $N$, and $P$ are $A$-modules ($A$ commutative, unital), I would like to see why $$(M\otimes N)\otimes P\stackrel{f}{\to}M\otimes N\otimes P$$ for $f((x\otimes y)\otimes z)=x\otimes y\otimes z$ is a homomorphism.

I think the map $(x,y)\mapsto x\otimes y\otimes z$ is bilinear, since for example, $$ (x+x',y)\mapsto (x+x')\otimes y\otimes z=[(x+x')\otimes y]\otimes z=[(x\otimes y)+(x'\otimes y)]\otimes z=(x\otimes y\otimes z)+(x'\otimes y\otimes z). $$ I just fear I might be unfairly assuming some associativity/distributive property that I'm attempting to prove.

Moreover, I've read that since bilinearity in $x$ and $y$ induces a homomorphism $f_z\colon M\otimes N\to M\otimes N\otimes P$ by $f_z(x\otimes y)=x\otimes y\otimes z$.

Would someone be kind enough to maybe explicitly show the calculation that $f_z$ is a module homomorphism? I think I will be more comfortable verifying other such isomorphisms between tensor products if I'm aware of what operations are allowable and those that are not. Thanks.

For instance, I was trying to figure out the image of say $(x\otimes y)+(x'\otimes y')$ under $f_z$, but I'm not fully confident how to compose this properly in $M\otimes N$ to find the right image.

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What you write certainly seems to assume what you want to prove, since you are conflating $a\otimes b\otimes c$ with $(a\otimes b)\otimes c$ several times. But more importantly: what is your definition of "$M\otimes N\otimes P$"? $M\otimes N$ has the universal property relative to bilinear maps; are you defining $M\otimes N\otimes P$ as the map with the corresponding universal property for multilinear maps? –  Arturo Magidin Jan 23 '12 at 21:38
    
@ArturoMagidin I'm not quite sure, I'm getting this from Proposition 2.14 of Atiyah and MacDonald. I think they would defined $M\otimes N\otimes P$ as the "multi-tensor product" generated by all products $x\otimes y\otimes z$, $(x\in M, y\in N,z\in P)$, but I find this troublesome since I don't fully understand what a product $x\otimes y\otimes z$ is, without parentheses, at least, since there is no assumption of associativity, as far as I can see. –  coffey Jan 23 '12 at 21:42
    
Have a look at the first part of Cartan and Eilenberg's Homological Algebra where they discuss associativity of the tensor product. –  mt_ Jan 23 '12 at 22:55
    
@coffey: My point is that in your development, you go from $(x+x')\otimes y\otimes z$ to $[(x+x')\otimes y]\otimes z$. Here, you are already assuming that there is a correspondence between tensors $a\otimes b\otimes c$ of $M\otimes N\otimes P$ and tensors $(a\otimes b)\otimes c$ of $(M\otimes N)\otimes P$, so you cannot use the argument to establish that correspondence. In order to establish that $(x+x')\otimes y\otimes z = (x\otimes y\otimes z) + (x'\otimes y\otimes z)$, you need to somehow use the precise description of $M\otimes N\otimes P$. –  Arturo Magidin Jan 24 '12 at 1:54
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