Does completing a normed space commute with taking quotients? Let $X$ be a normed vector space and $Y \subset X$ a closed subspace. We consider the quotient $X / Y$ and equip it with the quotient norm. Then we may form the completion $\overline{X / Y}$.
We compare $\overline{X / Y}$ to the following space: denote by $\overline{X}$ the completion of $X$ and by $\overline{Y} \subset \overline{X}$ the closure of $Y \subset \overline{X}$. Now we form the quotient $\overline{X} / \overline{Y}$ with the quotient norm.

Are $\overline{X / Y}$ and $\overline{X} / \overline{Y}$ (naturally) isomorphic, i.e., is there a linear, bijective isometry between them?

 A: Yes. We have the canonical map $\varphi = \overline{\pi} \circ \iota \colon X \to \overline{X}/\overline{Y}$ which factors through $X/Y$, and the induced map $\overline{\varphi} \colon X/Y\to \overline{X}/\overline{Y}$ is an isometry, since for every $\xi \in \overline{X}$ we have $\operatorname{dist}(\xi,Y) = \operatorname{dist}(\xi,\overline{Y})$. In particular, for $\xi \in X$, we have 
$$\lVert\pi(\xi)\rVert_{X/Y} = \operatorname{dist}(\xi,Y) = \operatorname{dist}(\xi,\overline{Y}) = \lVert \overline{\pi}(\iota(\xi))\rVert_{\overline{X}/\overline{Y}} = \lVert \overline{\varphi}(\pi(\xi))\rVert_{\overline{X}/\overline{Y}}.$$
Since $X$ is dense in $\overline{X}$, the image of $\overline{\varphi}$ is dense:
$$\overline{\overline{\varphi}(X/Y)} = \overline{\varphi(X)} = \overline{\pi(\iota(X))}\supset \pi\left(\overline{\iota(X)}\right) = \pi\left(\overline{X}\right) = \overline{X}/\overline{Y}.$$
The isometry $\overline{\varphi}$ extends to an isometry $\tilde{\varphi}\colon \overline{X/Y} \to \overline{X}/\overline{Y}$. Since $\tilde{\varphi}$ is an isometry, $\tilde{\varphi}\left(\overline{X/Y}\right)$ is a complete subspace of $\overline{X}/\overline{Y}$, hence closed, and since it contains $\overline{\varphi}(X/Y)$, it is a dense subspace. A closed dense subspace must be the entire space, so $\tilde{\varphi}$ is a bijection.
