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(Note: I'm using the word "natural" to mean "without the need to choose a basis." I'm aware that there is a precise category-theoretic meaning of this word, but I don't have great intuition for it yet and am hoping, perhaps naively, it's not necessary to understand the following.)

There exists a natural injection $V\rightarrow (V^{*})^{*}$ defined by sending $v\in V$ to a functional $\mu_{v}$ on $V^{*}$ such that $\mu_{v}(f)=f(v)$ for all $f\in V^{*}$. When $V$ is finite dimensional, this map is an isomorphism by comparing dimensions, so there is also an injection $(V^{*})^{*}\rightarrow V$. Is there a "natural" (again, in this context I understand this to mean basis-free) way to write down this injection, other than as simply the reverse of the first one?

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    there is, however, a natural map from the triple dual down to the first dual, which is an isomorphism for finite-dimensional spaces. So if you are willing to use the fact that every finite-dimensional space is the dual of some other space, then you do at least get a natural map, but it won't be an injection unless one is in the finite-dimensional setting2012-01-30

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For the sake of having an answer: no. Any good definition of "natural" would imply that this map also existed for infinite-dimensional vector spaces, which it doesn't. You shouldn't be able to do any better than "the inverse, when it exists, of the natural map $V \to (V^{\ast})^{\ast}$."