Dual of the algebra of endomorphisms Let $V$ be a complex vector space of arbitrary dimension. I'm supposed to prove that the isomorphism

$\text{End}(V) \to \text{End}(V)^*$ is canonical.

To start with, I am not even sure what this map should look like.
I know how to identify elements of a vector space with its dual, but this identification is non-canonical.
I'm not sure how replacing a vector space with its algebra of endomorphisms makes this identification canonical.
One other approach I can think of is using the isomorphism $\text{End}(V) \cong V \otimes V^*$ and working from there, but I still cannot figure out a map, let alone how it would be canonical.
 A: They won't even have the same dimension unless $V$ is finite-dimensional. If $V$ is finite-dimensional, there is a canonical isomorphism, which abstractly is given by the composite of the sequence of isomorphisms
$$\text{End}(V) \cong V \otimes V^{\ast} \cong (V \otimes V^{\ast})^{\ast} \cong \text{End}(V)^{\ast}$$
(where we use in the middle the natural isomorphism $(V \otimes W)^{\ast} \cong W^{\ast} \otimes V^{\ast}$ as well as using the double dual isomorphism; this is the most important step), and which concretely is given by
$$\text{End}(V) \ni X \mapsto \left( Y \mapsto \text{tr}(XY) \right) \in \text{End}(V)^{\ast}.$$
It's not entirely clear that these give the same map but you can verify it e.g. by working everything out in terms of a basis of $V$ and the corresponding dual basis of $V^{\ast}$. In terms of such a basis $\{ v_i \}$ and its dual basis $\{ v_i^{\ast} \}$ the first sequence of isomorphisms goes
$$X \mapsto \sum X_{ij} v_i \otimes v_j^{\ast} \mapsto \left( \sum Y_{ij} v_i \otimes v_j^{\ast} \mapsto \sum X_{ij} Y_{ji} \right) \mapsto \left( Y \mapsto \text{tr}(XY) \right)$$
which is the second isomorphism.
It's worth saying precisely that "canonical" here means, among other things, that every map I've written down is $GL(V)$-equivariant.
