Can a continuous map between a Banach space and a non-Banach space be bijective? Let $\mathbb{X}$ be a normed space that is complete and $\mathbb{Y}$ be another normed space which is not complete. Then can a bounded linear map $A:\mathbb{X} \to \mathbb{Y}$ be bijective or not?
 A: Edit: In fact there's a very simple theorem here that gives the whole truth: Given a bounded linear bijection $T:X\to Y$, where $X$ is complete, $Y$ is complete if and only if $T^{-1}$ is bounded. (If $Y$ is complete the open mapping theorem shows that $T^{-1}$ is bounded. On the other hand if $T^{-1}$ is bounded it's trivial to show that $Y$ is complete: A Cauchy sequence in $Y$ comes from a Cauchy sequence in $X$, which converges...)

Original:
Yes, it's possible. This surprises me; I thought the answer was no. The reason I thought the answer was no was something like this:
Let's agree that an isomorphism in the present context is a bounded linear bijection whose inverse is also bounded. Now (i) a bounded linear bijection between Banach spaces must be an isomorphism, (ii) if $X$ and $Y$ are isomorphic normed spaces and $X$ is complete then $Y$ is complete. Of course I never thought that was actually a proof here; all it proves is that $Y$ is complete if $Y$ is complete. But those facts in my head made me think the answer was no. 
Anyway, here's an example. Let $X=\ell^2$, the usual space of square-summable sequences. Define $T:X\to X$ by $$Tx=(x_1,x_2/2,x_3/3,\dots).$$
Then $T$ is certainly bounded and injective. Now let $Y=T(X)$, and give $Y$ the norm it inherits from $\ell^2$. Regard $T$ as a map from $X$ to $Y$. It's still bounded and injective, and now it's surjective.
So $T:X\to Y$ is a bounded linear bijection. And $Y$ is not complete. (Proof: If $Y$ were complete then the open mapping theorem would show that $T^{-1}:Y\to X$ was bounded, but $T^{-1}$ is certainly not bounded.)
A: $\newcommand{nrm}[1]{\left\lVert{#1}\right\rVert}$ Long story short: yes, and it happens quite often.
For instance, let $I=(0,1)$.
Consider the map \begin{align}\psi:W^{1,p}(I)&\hookrightarrow L^p(I)\\u&\mapsto u\end{align}
Since $\nrm{u}_{W^{1,p}}=\nrm{u'}_p+\nrm{u}_p$, it holds $\nrm\psi\le1$.
But $\psi\left(W^{1,p}(I)\right)=F$ contains $C^\infty_c(I)$, therefore it is dense in $L^p(I)$. Obviously, $F\ne L^p(I)$ because all functions in $F$ have finite $\operatorname{supess}$.
Hence $(F,\nrm{\bullet}_p)$ is not Banach and $\psi:W^{1,p}(I)\to F$ is continuous and bijective.
