given u in U, it is enough to show that φ(u) is in Ann(Ann(U)) (this proof only works in finite-dimensional vector spaces, by the way, as in general, U is not isomorphic to U**).

but if f is in Ann(U), this means f(u) = 0 for all u in U. hence for any such f, φ(u)(f) = f(u) = 0, that is φ(u) is contained in Ann(Ann(U)), which means that φ(U) = U** is contained in Ann(Ann(U)).

but clearly Ann(Ann(U)) is a subset of U**, so the two sets are equal (the finite-dimensionality comes in by way of asserting φ is ONTO U**).

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now suppose we take G in Ann(Ann(U)). since Ann(Ann(U)) = U**, and φ is an isomorphism, φ^{-1}is well-defined. letting u = φ^{-1}(G), we have:

φ(u) = φ(φ^{-1}(G)) = id_{U**}(G) = G.