Let $U,V,W$ be vector spaces over a field $K$ and $\varphi: U\to V,$ $\psi: V\to W$ linear maps. Given that
$\text{ker}(\psi)=\text{im}(\varphi)$ show that $\ker(\varphi^{tr})=\operatorname{im}(\psi^{tr})$
I tried proving this by double inclusion.
So far I have:
$"\subseteq"$
Let $x \in \text{im}(\psi^{tr}),$ then I know there exists $\alpha \in W^*$ such that $\alpha \circ \psi=x$ and $\varphi^{tr}(\alpha \circ \psi)=\alpha \circ \psi \circ \varphi$ is the zero mapping as $\ker(\psi)=\operatorname{im}(\varphi)$.
Can someone help me with the other inclusion?
Best Answer
If $\alpha \in \ker(\varphi^{tr})$, then $\alpha\circ\varphi$ is the zero map. By $\ker(\psi)=Im(\varphi)$, it follows that $\alpha|_{\ker(\psi)}=0$, it means that $\alpha \in (\ker \psi)^0$ (the annihilator of $\psi$) from This Question it follows that $\alpha \in Im(\psi^{tr})$.