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


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!


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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