If $U$ and $V$ are finite-dimensional vector spaces then $U^*\otimes V^* \approx (U \otimes V)^*$ via the isomorphism $\tau: U^*\otimes V^* \to(U \otimes V)^*$ given by $\tau(f \otimes g)(u \otimes v) = f(u)g(v)$.

And more generally the linear transformation $\theta: \mathcal{L}(U,U') \otimes \mathcal{L}(V,V') \to \mathcal{L}(U \otimes V, U' \otimes V')$ defined by $\theta(\tau \otimes \sigma) = \tau \odot \sigma$ where $(\tau \otimes \sigma)(u \otimes v) = \tau(u) \otimes \sigma(v)$ is an injective linear transformation and is an isomorphism if all vector spaces are finite-dimensional.

In the proof of these what exactly goes wrong for infinite dimensional vector spaces so that the linear transformation is not surjective.

An example showing non-surjective would also be appreciated.


Let $V$ be an infinite dimensional vector space over $k$. Let $W$ be the image of

$$\mathcal L(V, k)\otimes \mathcal L(k,V) \to \mathcal L(V \otimes k, k\otimes V).$$

Let us identify these objects and this map with

$$V^\vee \otimes V \to \mathcal L(V, V).$$

Suppose that $f : V \to V$ is in $W$. Show that $f(V)$ is finite-dimensional. In particular, the identity map of $V$ is not in $W$.

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