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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. –  Matt Pressland Oct 9 '12 at 14:49
    
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 $V$. –  chango Oct 9 '12 at 16:19
1  
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. –  Matt Pressland Oct 9 '12 at 16:23
    
I think I have figured it out. I will write this when I come back from the pub :) –  chango Oct 10 '12 at 19:34

1 Answer 1

Ok, so one pint too many but here I am coffee at hand. The only thing I use for the proof is the fact that $(\delta_s)_{s \in S}$ is a basis for $\mathcal{F}_{00}(S;\mathbb{K})$. It is pretty straightforward really...

PROOF:

Define $Ae_s = \alpha(s)$ for all $s \in S$. Let $v\in V$. Then by statement A we know $\exists! \: f \in \mathcal{F}_{00}(S;\mathbb{K})$ such that $$\Delta^{-1} f = v. $$ Since $(\delta_s)_{s \in S}$ is a basis for $\mathcal{F}_{00}(S;\mathbb{K})$ we can write $$ f = \sum_{s \in S} f(s) \delta_s, $$ $$\implies v = \Delta^{-1}(\sum_{s \in S} f(s) \delta_s ) = \sum_{s \in S} f(s) \Delta^{-1} \delta_s = \sum_{s \in S} f(s) e_s, $$ $$\text{linearity of A } \implies Av = \sum_{s \in S} f(s) Ae_s = \sum_{s \in S} f(s) \alpha(s). $$

Hence we have shown existence and uniqueness.

$\square$

Note: that the above sum is finite since $f$ only takes on finitely many non-zero values.

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