$(H_1 \otimes H_2) \otimes H_3 \cong H_1 \otimes H_2 \otimes H_3 \cong H_1 \otimes (H_2 \otimes H_3)$ Let $H_1, H_2, H_3$ be Hilbert spaces. In this post, $\odot$ denotes an algebraic tensor product and $\otimes$ a tensor product of Hilbert spaces (i.e. a completion of an algebraic tensor product w.r.t. the canonical inner product). I want to show that $$(H_1 \otimes H_2) \otimes H_3 \cong H_1 \otimes H_2 \otimes H_3 \cong H_1 \otimes (H_2 \otimes H_3)$$
if this is even true?
I tried the following:
The canonical map $$\psi: H_1 \odot H_2 \odot H_3 \to (H_1 \odot H_2) \odot H_3$$
is easily checked to be isometric. Hence, $\psi$ extends to an isometry $$\psi: H_1 \otimes H_2 \otimes H_3 \to (H_1\otimes H_2) \otimes H_3$$
and since $(H_1 \odot H_2)\odot H_3 $ is dense in $(H_1 \otimes H_2) \otimes H_3$, $\psi$ is surjective. Hence, we have an isometric isomorphism, as desired.
The other isomorphism can be shown similarly.
Is the above correct?
 A: Once you have an isometry between two pre-Hilbert spaces $A$ and $B$, you know that the resulting Hilbert spaces $\overline{A}$ and $\overline{B}$ are isometric as well, since your isometry will canonically extend to the completions.
However, the Hilbert space $(H_1\otimes H_2)\otimes H_3$ is by definition the completion of the pre-Hilbert space $(H_1\otimes H_1)\odot H_3$, not of $(H_1\odot H_2)\odot H_3$.
Hence, what you need to check is that going from $(H_1\odot H_2)\odot H_3$ to $(H_1\otimes H_2)\odot H_3$ by completing the first factor and then to $(H_1\otimes H_2)\otimes H_3$ by completing the result yields a Hilbert space that is canonically isometric to what you get from directly completing the pre-Hilbert space $(H_1\odot H_3)\odot H_3$.
So, in your attempt, the extension of $\psi$ to completions should be
$$
\overline \psi \colon \underbrace{H_1\otimes H_2\otimes H_3}_{= \overline{H_1\odot H_2\odot H_3}} \stackrel{\cong}\longrightarrow \overline{(H_1\odot H_2)\odot H_3}
$$
and you are missing an isometry
$$
\overline{(H_1\odot H_2)\odot H_3} \cong \underbrace{(H_1\otimes H_2)\otimes H_3}_{=\overline{\left(\overline{H_1\odot H_2}\right)\odot H_3}}.
$$

See Proposition 2.6.5 in R.V. Kadison, J. R. Ringrose: Fundamentals of the Theory of Operator Algebras (1983) for a general proof of
$$H_1\otimes \dots \otimes H_{n+m} \cong (H_1\otimes \dots\otimes H_n)\otimes(H_{n+1}\otimes\dots\otimes H_{n+m})$$
using a universal property of tensor products of Hilbert spaces.
