Example for $\text {End}(V \otimes W) $ not isomorphic to $\text{End}(V) \otimes \text{End}(W)$ A couple of posts relate to this topic and indicate that when the spaces are finite dimensional there is a canonical isomorphism, while in the infinite dimensional case there is only an injection $\mathrm{End}(V) \otimes \mathrm{End}(W) \to \mathrm{End}(V \otimes W)$.
Do we have $End(V \otimes V) = End(V) \otimes End(V)$?
Operators on a Tensor Product Space
https://mathoverflow.net/q/72013
So, is there an example that demonstrates this ?
I considered infinite dimensional spaces over the reals with dimension $c$ and concluded that

*

*With $\dim(V) = \dim(W) = c$

*$\dim (\text{End} (V)) = c^c$; so $\dim (\text{End} (V) \otimes \text{End}(W)) = c^c.c^c = c^c$;

*$\dim (V \otimes W) = c.c = c$

*$\dim(\text {End}(V \otimes W) ) = c^c = \dim (\text{End} (V) \otimes \text{End}(W))$
So, if my (cardinal) arithmetic is correct they are isomorphic in this case.

In response to the first two comments, I see that the fact that the given canonical injection  is not an isomorphism doesn't preclude the spaces being isomorphic. So my question is are they always isomorphic in the infinite dimensional case or is there an example to show they are not ?
 A: I'll assume the axiom of choice. I don't know what weirdness may happen otherwise here.
Let's suppose that $\kappa = \dim V \leqslant \lambda = \dim W$, and $\lambda$ is an infinite cardinal.
If $\kappa = 0$, then $\operatorname{End} (V\otimes W) = \operatorname{End} \{0\} = \{0\} = (\operatorname{End} \{0\}) \otimes (\operatorname{End} W)$. And if $\kappa > 0$, then we have
$$\dim (V \otimes W) = \kappa\cdot \lambda = \lambda = \dim W,$$
so it follows that $V \otimes W \cong W$, and consequently
$$\operatorname{End} (V \otimes W) \cong \operatorname{End} W.$$
Thus, with the canonical injection $(\operatorname{End} V) \otimes (\operatorname{End} W) \to \operatorname{End} (V \otimes W)$ we deduce
$$\dim (\operatorname{End} W) \leqslant \dim \bigl((\operatorname{End} V) \otimes (\operatorname{End} W)\bigr) \leqslant \dim \operatorname{End} (V \otimes W) = \dim (\operatorname{End} W),$$
whence there always is an abstract isomorphism between $(\operatorname{End} V) \otimes (\operatorname{End} W)$ and $\operatorname{End} (V \otimes W)$, whatever the dimensions of the spaces. But when both spaces are infinite-dimensional, the canonical map is not an isomorphism, and that's what people are interested in.
