Let $X$ be a normed space, $M$ a closed subspace, and a quotient of $X$ over $M$ defined as an ordered pair $(Q, \pi)$ such that
- $Q$ is a normed linear space
- $\pi$ is a bounded linear function from $X$ to $Q$
- $Ker(\pi)$ is contained in $M$
- For any other $(\hat{Q},\hat{\pi})$ that satisfies the above bullets, there exists a unique bounded linear function $\xi:Q\rightarrow\hat{Q}$ such that $$\hat{\pi} = \xi\circ\pi$$
Prove that there exist bounded linear functions $Q\rightarrow\hat{Q}$ and $\hat{Q}\rightarrow Q$ that are inverses of each other.
It's easy enough to see, by composition, that we have linear functions that equal the identity when restricted to $\pi(X)$ and $\hat{\pi}(X)$. Hahn-Banach says that this bounded linear function is extendable to one of the same norm (i.e., one) on the whole of $Q$ (or $\hat{Q}$). But that itself does not guarantee that the function on the whole space $Q$ ($\hat{Q}$) is the identity ... does it?
Let $\xi \colon Q \to \hat Q$ and $\hat\xi \colon \hat Q \to Q$ the unique bounded linear maps such that $$ \hat pi = \xi \pi, \qquad \pi = \hat\xi \hat\pi $$ which exist by the universal property.
Now consider the two maps $\def\i{\operatorname{id}}\i, \hat\xi\xi \colon Q\to Q$. We have $$ \i\pi = \pi, \qquad \hat\xi \xi \pi = \hat\xi\hat\pi = \xi $$ But, as $(Q,\pi)$ is a quotient, there is a unique bounded linear map $f \colon Q\to Q$ such that $$ f\pi = \pi $$ As $\i$ and $\hat\xi \xi$ both have this property, we have $\i = \hat\xi \xi$.
Along the same line of thought, exchanging the roles of $X$ and $\hat X$, $\i = \xi\hat\xi$. So $\xi$ and $\hat\xi$ are mutual inverses.