Tensor product of maps between Hilbert spaces Let $H_1,H_2, K_1,K_2$ be Hilbert spaces. Let $S_1: H_1 \to K_1$ and $S_2: H_2 \to K_2$ be bounded linear maps. Is it true that there exists a bounded linear map
$$S_1 \otimes S_2: H_1 \otimes H_2 \to K_1 \otimes K_2$$ such that $(S_1 \otimes S_2) (\xi_1 \otimes \xi_2) = (S_1 \xi_1) \otimes (S_2 \xi_2)$? Here the tensor products are not algebraic tensor products! Rather, they are completions of the algebraic tensor product.
I know this result is true when $H_1 = K_1$ and $H_2 = K_2$, but is this true as well?
 A: The approach  based on decomposing $S_1$ and $S_2$ as  linear combinations of unitary operators  does prove  that $S_1\otimes S_2$ is
bounded but, as far as I can remember, does not directly give the sharp estimate
$$
  \|S_1\otimes S_2\|\leq   \|S_1\|\|S_2\|,
  $$
since the coefficients in the aforementioned linear combinations have little relationship with the norm of the
operators involved.
Here is a different approach,  completely avoiding  unitaries, and which gives the desired norm estimate.
Lemma 1.  If $T:H_1\to H_2$ is a bounded linear operator between  Hilbert spaces $H_1$ and $H_2$, and given $x_1,x_2,\ldots ,x_n\in H_1$, let $A$ and $B$ be the $n\times n$
matrices defined  by
$$
  a_{ij} = \langle T(x_i),T(x_j)\rangle ,
  \quad \text{and} \quad
  b_{ij} = \langle x_i,x_j\rangle .
  $$
Then $A\leq  \|T\|^2 B$.
Proof.  Given scalars  $\lambda _1,\lambda _2,\ldots ,\lambda _n$, we have
$$
  \sum_{i,j}\lambda _i\bar\lambda _ja_{ij} =
  \sum_{i,j}\lambda _i\bar\lambda _j\langle T(x_i),T(x_j)\rangle  =
  \sum_{i,j}\langle T(\lambda _ix_i),T(\lambda _jx_j)\rangle  = $$$$ =
  \left\langle \sum_iT(\lambda _ix_i),\sum_jT(\lambda _jx_j)\right\rangle  =
  \left\|T\left(\sum_i \lambda _ix_i\right)\right\|^2 \leq  $$$$ \leq
  \|T\|^2\left\|\sum_i \lambda _ix_i\right\|^2 =
  \|T\|^2  \sum_{i,j}\langle \lambda _ix_i,\lambda _jx_j\rangle  =  $$$$ =
  \|T\|^2   \sum_{i,j}\lambda _i\bar\lambda _j\langle x_i,x_j\rangle  =
  \|T\|^2    \sum_{i,j}\lambda _i\bar\lambda _jb_{ij}.
  $$
QED
Lemma 2.  If $T:H_1\to H_2$ is a bounded linear operator between  Hilbert spaces $H_1$ and $H_2$, and $K$ is another Hilbert
space, then there is a unique  linear operator
$$
  T\otimes I:H_1\otimes K\to H_2\otimes K,
  $$
such that
$$
  T(x\otimes y) = T(x)\otimes y, \quad \forall x\in  H_1, \quad \forall y\in  K.
  $$
Moreover $\|T\otimes I\|\leq \|T\|$.
Proof.  By replacing $T$ with $T/\|T\|$, we may assume that $\|T\|=1$.
The crucial step in the proof is then showing the inequality
$$
  \left\| \sum_i T(x_i)\otimes y_i \right\| \leq    \left\| \sum_i x_i\otimes y_i \right\|,
  $$
whenever $x_1,x_2,\ldots ,x_n\in H_1$, and $y_1,y_2,\ldots ,y_n\in K$.
In this case,
observe that the matrices $A$ and $B$ defined in Lemma (1)  satisfy $A\leq  B$, whence $B-A\geq 0$,
so we may find an $n\times n$ scalar matrix $C$ such that
$B-A=CC^*$.  Therefore, for every $i$ and $j$ one has that
$$
  \langle x_i, x_j\rangle  = \langle T(x_i), T(x_j)\rangle  +  \sum_k c_{ik}\bar c_{j k}.
  $$
Multiplying by   $\langle y_i, y_j\rangle $ and summing on $i$ and $j$, we get
$$
  \sum_{i, j}\langle x_i, x_j\rangle   \langle y_i, y_j\rangle  = \sum_{i, j}\langle T(x_i), T(x_j)\rangle \langle y_i, y_j\rangle  +  \sum_{i, j}\sum_k c_{ik}\bar c_{jk}\langle y_i, y_j\rangle ,
  $$
which means that
$$
  \left\|\sum_i x_i\otimes y_i \right\|^2 = \left\|\sum_i T(x_i)\otimes y_i \right\|^2 +  \sum_{i, j}\sum_k c_{ik}\bar c_{j k}\langle y_i, y_j\rangle ,
  $$
so it suffices to show that the last sum above is nonnegative.  But this is easy because
$$
  \sum_{i, j}\sum_k c_{ik}\bar c_{j k}\langle y_i, y_j\rangle  =
  \sum_k \left\langle \sum_i c_{ik}y_i,\sum_jc_{j k}y_j\right\rangle  =
  \sum_k \left\|\sum_i c_{ik}y_i\right\|^2 \geq 0.
  $$
QED
Adopting the notation used by the OP, we may now easily prove  that $S_1\otimes S_2$ is bounded, and indeed that
$\|S_1\otimes S_2\|\leq \|S_1\|\|S_2\|$, as follows
$$
  \|S_1\otimes S_2\| =   \|(S_1\otimes I)(I\otimes S_2)\| \leq  \|S_1\otimes I\|\|I\otimes S_2\| \leq   \|S_1\|\|S_2\|.
  $$
A: Embed each of your spaces as a closed subspace of a bigger space, say $H_1\subset\tilde H_1$, etc., such that $$\dim\tilde H_1=\dim\tilde K_1,\qquad \dim\tilde H_2=\dim\tilde K_2.$$ You can trivially extend $S_1$ and $S_2$ to operators $\tilde S_1,\tilde S_2$ by making them $0$ on $H_1^\perp$ and $H_2^\perp$ respectively.  Let $U:\tilde H_1\to\tilde K_1$, $V:\tilde H_2\to\tilde K_2$ be unitaries. Now you can use the existence of a bounded linear map
$$
U^*\tilde S_1\otimes V^*\tilde S_2:\tilde H_1\otimes \tilde H_2\to \tilde H_1\otimes \tilde H_2
$$
that acts on elementary tensors as you want. Now note that for unitaries, it is trivial to check that $U\otimes V$ is isometric on sums of elementary tensors, which means that $U\otimes V$ is a well-defined unitary on the tensor product. So we can define
$$
S_1\otimes S_2=(U\otimes V)(U^*\tilde S_1\otimes V^*\tilde S_2)|_{H_1\otimes H_2}.
$$
This map is bounded, being a composition of bounded operators, and
$$
(S_1\otimes S_2)(\xi_1\otimes \xi_2)=(U\otimes V)(U^*\tilde S_1\xi_1\otimes V^*\tilde S_2\xi_2)
=S_1\xi_1\otimes S_2\xi_2. 
$$
