Show that $Y$ is complete. Let $X$, $Y$ be normed linear spaces and $T \in B(X,Y)$. Show that if $X$ is complete and $T$ is an onto open map, then $Y$ is complete.
where, $B(X,Y)$ are bounded linear functions from $ X$ to $Y$.
Can anyone give me some hints?
 A: 
Let $T \in \mathcal{L}(X,Y)$ be open and surjective, then there is a constant $C$ such that $ \forall y \in Y$ there is an $x \in T^{-1}(y)$ such that $$\|x\|_X \le C\|y\|_Y.$$

Proof: $T(B_X(0,1))$ is open, then it contains an open ball centered at $0 \in Y$: $B_Y(0,\epsilon) \subset T(B_X(0,1))$. Let $y \in T(X) \setminus \{0\}$. Then $$z := \frac{\epsilon}{2}\frac{y}{\|y\|_Y} \in B_Y(0,\epsilon) \subset T_X(B(0,1)),$$ and thus there exists $x_0 \in B_X(0,1)$ with $T(x_0) = z.$ Note that by linearity of $T$, $$T(x_0) = z = \frac{\epsilon}{2}\frac{y}{\|y\|_Y} \Longrightarrow T\Big(\frac{2\|y\|_Y}{\epsilon}x_0\Big) = y.$$ Set $x = \frac{2\|y\|_Y}{\epsilon}x_0.$ Then $T(x) = y$ and, since $x_0 \in B_X(0,1)$, we have $$\|x\|_X = \frac{2\|y\|_Y}{\epsilon}\|x_0\|_X \le \frac{2\|y\|_Y}{\epsilon}.$$ Hence the claim holds with $C = \frac{2}{\epsilon}.$ $\blacksquare$

Now take a Cauchy sequence in $Y$ and pull it back to the complete space $X$ via $T^{-1}$, making sure to pick $x \in T^{-1}(y)$ as in the above proposition.
