Does the "2nd isomorphism theorem" hold for Banach spaces, when one of the subspaces is finite-dimensional? Let $X$ be a Banach space. Let $Y, F \subseteq X$ be subspaces with $Y$ closed and $F$ finite-dimensional. It follows that $F + Y$ is also a closed subspace of $X$. 
Question: Is there an isometric isomorphism of finite-dimensional Banach spaces between $\frac{F+Y}{Y}$ and $\frac{F}{F \cap Y}$?
 A: No, or at least the canonical bijection is typically not an isometry.  For instance, let $X=\mathbb{R}^2$ with the Euclidean norm, let $F$ be the span of $(1,1)$, and let $Y$ be the span of $(1,0)$.  Then $F\cap Y=0$, so $\frac{F}{F\cap Y}=F$.  But the canonical bijection sends the element $(1,1)\in F$ to the coset $(1,1)+Y=(0,1)+Y$ in $\frac{F+Y}{Y}$, and $\|(1,1)\|=\sqrt{2}$ but $\|(0,1)+Y\|=1$.
(In this example, $\frac{F}{F\cap Y}$ and $\frac{F+Y}{Y}$ happen to be isometrically isomorphic, but not by the natural map.  Probably you can get a similar example where they are not isometrically isomorphic by any map (say, take a similar example where the quotient spaces end up being 2-dimensional and you have a less symmetrical norm than the Euclidean norm), but proving that no map can be an isometry is probably kind of a mess.)
However, the canonical bijection is always an isomorphism of topological vector spaces.  Indeed, the canonical bijection is norm-decreasing as a map $\frac{F}{F\cap Y}\to\frac{F+Y}{Y}$, since given $f\in F$, the distance from $f$ to $F\cap Y$ cannot be smaller than the distance from $f$ to $Y$.  So the canonical bijection is a bounded linear bijection $\frac{F}{F\cap Y}\to\frac{F+Y}{Y}$, which is hence a topological isomorphism by the open mapping theorem.
