Injectivity/surjectivity duality

Yet again I’m unable to sleep, so here’s something else I was thinking about. Suppose we fix a field \mathbb{F} and let \mathsf{V} and \mathsf{W} be two vector spaces over \mathbb{F}.

Suppose \varphi : \mathsf{V} \to \mathsf{W} is an injective linear map. Let \varphi^* : \mathsf{W}^* \to \mathsf{V}^* be the induced dual (transpose) map. Recall this is given by \varphi^*(f) = f \circ \varphi. I claim that \varphi^* is surjective.

To prove this, let’s pick an arbitrary \psi \in \mathsf{V}^* and attempt to construct a preimage of this thing under \varphi^*. That is, we want to find an f \in \mathsf{W}^* such that \varphi^*(f) = \psi. Let a basis \{ v_\alpha \}_{\alpha \in \Lambda} for \mathsf{V} be given. Observe that the injectivity of \varphi tells us that \{ \varphi(v_\alpha) \}_{\alpha \in \Lambda} will form a basis for the image of \varphi. OK. Now, define the numbers \zeta_\alpha := \psi(v_\alpha). To form our function f, we need do nothing more than define it on a basis: put f(w) = \zeta_\alpha whenever w = \varphi(v_\alpha) for some \alpha \in \Lambda, and let f(w) = 0 otherwise (I’m abusing language here, but you know what I mean).

Clearly, we see that if v = \sum_{\alpha} \lambda_\alpha v_\alpha (note that this must be a finite sum by the definition of a basis) is an arbitrary vector in \mathsf{V}, then

\psi(v) = \sum_{\alpha} \lambda_\alpha \psi(v_\alpha) = \sum_{\alpha} \lambda_\alpha \zeta_\alpha

but

f(\varphi(v)) = \sum_{\alpha} \lambda_\alpha f(\varphi(v_\alpha)) = \sum_{\alpha} \lambda_\alpha \zeta_\alpha

by the way we defined f. Therefore f \circ \varphi = \psi and we’ve successfully finished our quest in demonstrating that \varphi^* is surjective.

I’m pretty sure that if, on the other hand, \varphi is assumed to be surjective, then we should be able to similarly conclude injectivity of \varphi^* (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.
This entry was posted in abstract algebra, linear algebra and tagged , , , , . Bookmark the permalink.

2 Responses to Injectivity/surjectivity duality

  1. mateusz says:

    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.

    • dx7hymcxnpq says:

      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 \{ \varphi(v_\alpha) \} to a basis \{ w_\beta \} for all of \mathsf{W} (this is valid even without finite-dimensionality), and then I’m defining f by defining it only on those basis vectors (which is enough to completely characterize any linear map).

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