Let $S$ be a subspace of a vector space $V$. Prove that for any subspace $\overline{K}$ of $V/S$ there exists a subspace K of V such that $S\leq K$ and $\overline{K}= K$.
I don't have a lot experience with quotient spaces. My idea for the proof is to use the first isomorphism theorem that indicates, given linear $\phi$ from V to W, that the spaces $V/ker(\phi)$ are isomorphic to $Im(\phi)$ and relate the images with the sets to find $K$, but I have no idea how to do it. Another idea was to use the projection and also relate the image to build the sets I need.
Psdt. $S\leq K$ indicates that $S$ is a subspace of $K$. (I did not know this representation)
u have a natural Linear surjection $\pi :V\to V/S, \pi(x)=x+S=[x]$ (associating to each vector a coset of the subspace $S$).U can prove the following fact :if $\overline{K}\leq V/S$,then $\pi^{-1}(\overline{K})\leq V \land S\subset \pi^{-1}(\overline{K})$ ,more precisely $\pi^{-1}(\overline{K})=S+\overline{K}$.Prove that $\pi(\pi^{-1}(\overline{K})=\overline{K}$,So $k=\pi^{-1}(\overline{K})$.In Conclusion,there is a one to one correspondance between subspaces of $V/S$ and subspaces of $V$ containing $S$: $$\pi:\{K\leq V \backslash S\subset K\}\to \{\overline{K}\leq V/S\} $$ $$\pi^{-1}:\{\overline{K}\leq V/S\} \to \{K\leq V \backslash S\subset K\}$$