Yet again I’m unable to sleep, so here’s something else I was thinking about. Suppose we fix a field and let and be two vector spaces over .

Suppose is an injective linear map. Let be the induced dual (transpose) map. Recall this is given by . I claim that is surjective.

To prove this, let’s pick an arbitrary and attempt to construct a preimage of this thing under . That is, we want to find an such that . Let a basis for be given. Observe that the injectivity of tells us that will form a basis for the image of . OK. Now, define the numbers . To form our function , we need do nothing more than define it on a basis: put whenever for some , and let otherwise (I’m abusing language here, but you know what I mean).

Clearly, we see that if (note that this must be a finite sum by the definition of a basis) is an arbitrary vector in , then

but

by the way we defined . Therefore and we’ve successfully finished our quest in demonstrating that is surjective.

I’m pretty sure that if, on the other hand, is assumed to be surjective, then we should be able to similarly conclude injectivity of (in this sense, injectivity and surjectivity are dual to one another). Good night!

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## About mlbaker

just another guy trying to make the diagrams commute.

not sure if i’m doing it right, but isn’t W* the space of all linear transformations? if so, shouldn’t f be linear? by your construction i don’t think it is. assuming f is linear, zeta_alpha_1 + zeta_alpha_2 = f(w_1) + f(w_2) = f(w_1 + w_2) = 0, since w_1 + w_2 is not phi(v_alpha) for some alpha.

Yeah true, that’s why I said I’m kind of abusing language. Implicitly, I’m more or less doing this: Invoke a basis extension theorem to extend to a basis for all of (this is valid even without finite-dimensionality), and then I’m defining by defining it only on those basis vectors (which is enough to completely characterize any linear map).