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I need help proving that statement A implies B:

STATEMENT A: $\exists!$ isomorphism $\Delta: V \to \mathcal{F}_{00}(S;\mathbb{K}) $ satisfying $\Delta e_s = \delta_s$ for all $s \in S$.

STATEMENT B: for any vector space $W$ over $\mathbb{K}$ and map $\alpha \in \mathcal{F}(S;W)$ there is a unique map $A \in L(V;W)$ s.t. $Ae_s=\alpha (s)$ for all $s \in S$.

NOTE: $(e_s)_s\in S$ is an indexed set in a vector space $V$ over $\mathbb{K}$ (real or complex field). $\mathcal{F}_{00}(S;\mathbb{K})$ denotes the set of maps from $S$ to $\mathbb{K}$ having finite support; whereas $\mathcal{F}(S;W)$ denotes the set of functions from $S$ to $W$.

Any hint, tip, etc will be appreciated. Thank you!

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    A good first step would seem to be to prove that $v=\sum_{s\in S}\Delta(v)(s)e_s$ for all $v\in V$. This gives you existence of $A$. I don't see uniqueness immediately (hence a comment rather than an answer), but I haven't yet used uniqueness of $\Delta$. This result seems morally the same as saying that linear maps are determined by their action on a basis in the case $S=\{1,\ldots,n\}$, so maybe some inspiration can be found in proofs of that.2012-10-09
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    Thanks. It is indeed a characterisation of a basis in the abstract setting. One should be able to prove that these two statements are equivalent to $(e_s)_{s \in S}$ being a basis for $V$.2012-10-09
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    This was in fact my first attempt at coming up with something useful to say, and it is indeed true that they are equivalent to $(e_s)_{s\in S}$ being a basis. However, I think it should be possible to prove what you want without ever uttering the word "basis", which is philosophically preferable in some ways.2012-10-09
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    I think I have figured it out. I will write this when I come back from the pub :)2012-10-10

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