## 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!

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).