Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics I / 3 / b / ii

For $u\in U$, the vector $Tu$ already belongs to $\mathrm{im}T$, soTherefore $\ker (ST)=T^{-1}(\ker S^{\prime })$. Since $\ker S^{\prime }=\ker S\cap \mathrm{im}T$, its intersection with $\mathrm{im}T$ is itself. Applying a with $X=\ker S^{\prime }$ yieldsThe inequality follows from the inclusion $\ker S^{\prime }\subseteq \ker S$ and monotonicity of dimension for vector subspaces.