# Associativity of Tensor Product

I have a doubt on the associativity of the tensor product. I know that the tensor product of vector spaces is an associative operation up to a linear isomorphism and I'm just trying to prove that.

My idea is: Let $V_1, \dots ,V_p$ be vector spaces over the same field $\mathbb{K}$ and let $k, r \in \mathbb{N}$ such that $1 < k < r < p$. Define then the spaces:

$$T_1=\bigotimes_{i=1}^{k}V_i$$

$$T_2=\bigotimes_{i=k+1}^{r}V_i$$

$$T_3=\bigotimes_{i=r+1}^{n}V_i$$

Then, if I proove that $T_1 \otimes (T_2 \otimes T_3) = (T_1 \otimes T_2) \otimes T_3$ I'll be proving the general case, so the entire proof is reduced to prove that the tensor product of three vector spaces is associative up to a linear isomorphism, so that we can use this with $T_1$, $T_2$ and $T_3$.

Is this idea correct ? This approach really gives a proof of this fact, or there's some problem approaching it this way ?

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For a general binary operation $*$ to be associative, is already defined as $$(x*y)*z=x*(y*z)$$ holds for all elements $x,y,z$. This basically enables us to write simply $x*y*z$ for $(x*y)*z=x*(y*z)$, i.e. naturally introduces a ternary operation. By induction, we can see that for any $n$, any parentheses put among $*$ of elements $x_1,x_2,..,x_n$ will result in the same element, i.e., naturally leading to the $n$-ary operation $x_1*x_2*...*x_n$.
This is almost the same for tensor product of vector spaces (or more generally, of modules over a commutative ring), we can just replace $=$ by $\cong$ everywhere above. That is, instead of equality, we have specific isomorphisms $\iota_{V_1,V_2,V_3}$ between $(V_1\otimes V_2)\otimes V_3$ and $V_1\otimes(V_2\otimes V_3)$, but the idea can still be applied: any parentheses put among the tensors of $V_1,V_2,..,V_n$ can be reordered using these specific isomorphisms to $(..((V_1\otimes V_2)\otimes V_3)...)\otimes V_n$. This proves that any two parentheses of a tensor are isomorphic to each other.
However, to ensure that, composing these corresponding $\iota$ isomorphisms, we always arrive to the same isomorphism, we need to check that for $n=4\$ (and then by MacLane's theorem it also follows for all $n>4$): $$((V_1\otimes V_2)\otimes V_3)\otimes V_4 \to (V_1\otimes(V_2\otimes V_3))\otimes V_4\to V_1\otimes((V_2\otimes V_3)\otimes V_4) \to V_1\otimes(V_2\otimes (V_3\otimes V_4)) \ = \\ ((V_1\otimes V_2)\otimes V_3)\otimes V_4 \to (V_1\otimes V_2)\otimes (V_3\otimes V_4)\to V_1\otimes(V_2\otimes (V_3\otimes V_4))$$ This is basically the concept of a monoidal category.
An alternative way is to explicitly introduce the $n$-ary tensor operation (just the same way as binary tensor is defined), and then we deal with canonical isomorphisms $V_1\otimes V_2\otimes...\otimes V_n\to V_1\otimes ..(V_i\otimes..\otimes V_j)..\otimes V_n$. This approach leads to the unbiased notion of monoidal category..