Given a Hilbert space $H$, let $S\_1(H)$ denote the space of trace-class operators on $H$, with the trace-class norm or Schatten 1-norm. That is
$$ \Vert T \Vert\_1 = \sum\_{j\geq 1} |s\_j| $$
where $(s\_1,s\_2,\dots)$ is the sequence of eigenvalues of the operator $|T|=(T^\*T)^{1/2})$ written in any order.


Let $H\_1$ and $H\_2$ be Hilbert spaces of dimention $n$ and $m$ respectively. Let $H\_1\otimes\_2 H\_2$ denote their Hilbert-space tensor product (hence, an inner product space of dimension $nm$).
Consider the map $$\varphi:S\_1(H\_1)\hat{\otimes}S\_1(H\_2)\to S\_1(H\_1\otimes\_2 H\_2),$$ which is defined by 
$$\varphi( A \otimes B) (\xi\_1\otimes\xi\_2) = (A\xi\_1 )\otimes (B\xi\_2) \quad(\xi\_1\in H\_1, \xi\_2\in H\_2) $$


Is it the case that $$\Vert\varphi^{-1}\Vert=\min\lbrace{m‎, ‎n}\rbrace ?$$
Here, $\hat{\otimes}$ denotes the projective tensor product of Banach spaces.



Comparing norms on tensor products of matrices
Revised answer:
===============


(Notation from p 61ff of the reference below).
Note that for trace class norm we have 
$\mathcal{B}(H\_1) = H\_1 \hat\otimes H\_1' = H\_1\hat \otimes H\_1$. For operator norm we have $\mathcal{B}(H\_1) = H\_1\hat{\hat\otimes} H\_1'= H\_1\hat{\hat\otimes} H\_1$ where the tensor product is now the inductive one (also called $\epsilon$-tensor product). Using this and finite dimensionality we then can write  

$\varphi:\mathcal{B}(H\_1)\hat{\otimes}\mathcal{B}(H\_2)\to\mathcal{B}(H\_1\hat{\otimes}H\_2)$
as the natural isomorphism
$$
(H\_1 \hat\otimes H\_1)\, \hat\otimes\, (H\_2 \hat\otimes H\_2) \to
(H\_1 \hat\otimes H\_2)\hat\otimes (H\_1 \hat\otimes H\_2)' =
(H\_1 \hat\otimes H\_2)\hat\otimes (H\_1 \hat{\hat\otimes} H\_2)
$$
which is $Id\,\hat\otimes\,j$ where $j:H\_1 \hat\otimes H\_2\to H\_1 \hat{\hat\otimes} H\_2$
is the embedding (here iso) of trace class norm into operator norm. The norm of $j$ is indeed $min\{\dim(H\_1,\dim H\_2)$, and the norm of $j^{-1}$ is its inverse, which can be seen by fixing bases and then writing any operator $A:H\_1\to H\_2$ as $U\_2\circ D\circ U\_1$ for isometries $U\_i$ and a diagonal operator $D$. 


Ah,
===


I see now that the question changed. Here is the version for the new question:
$$
(H\_1 \hat\otimes H\_1)\, \hat\otimes\, (H\_2 \hat\otimes H\_2) =
(H\_1 \hat\otimes H\_2)\, \hat\otimes\, (H\_1 \hat\otimes H\_2) 
\to
(H\_1 \otimes\_2 H\_2)\hat\otimes (H\_1 \otimes\_2 H\_2)' =
(H\_1 \otimes\_2 H\_2)\hat\otimes (H\_1 \otimes\_2 H\_2)
$$
which is $i\hat\otimes i$ where $i: H\_1 \hat\otimes H\_2 \to H\_1 \otimes\_2 H\_2$ is the natural iso. Its norm is the norm of the embedding $\ell^1\to \ell^2$ for the s-numbers. 


* Johann Cigler, Viktor Losert, Peter W. Michor: Banach modules and functors on categories of Banach spaces. Lecture Notes in Pure and Applied Mathematics 46, Marcel Dekker Inc., New York, Basel, (1979) [(pdf)](http://www.mat.univie.ac.at/~michor/banach_modules_and_functors.pdf)


