Banach spaces and surjective operators Let $X,Y$ be two Banach spaces and $T \in L(X,Y)$ be surjective. Then there exists a constant $C>0$ so that for every $y\in Y$ there exists a $x\in X$ with $Tx=y$ and that the following is true: $||x||_X \leq C||y||_Y$.
My thoughts: Surjection would mean that $\forall y\in Y \exists x \in X : Tx=y$. This is pretty much the first statement?
I'm not sure about the constant. Since $X,Y$ are banach spaces, I've thought of equivalent norms but I'm unsure how to prove this.
 A: This follows from the open mapping theorem: Let $K(x,r)$ denote the open ball with center $x$ and radius $r$. Since $T$ is surjective it is open. Thus there is some $r>0$ such that $K(0,r) \subseteq T(K(0,1))$. Let $y \in Y$ (w.l.o.g. $y \not=0$). Then $\frac{r}{2\|y\|}y \in K(0,r)$. Thus for some $z \in K(0,1)$ we have $Tz=\frac{r}{2\|y\|}y$, that is $y = T(\frac{2\|y\|}{r}z)$. Now $x:=\frac{2\|y\|}{r}z$ satisfies $Tx=y$ and $\|x\| \le \frac{2}{r}\|y\|$, so you can choose $C=\frac{2}{r}$.
A: Let $ \pi: X \to X/N(T)$ be defined by $\pi(x):=x+N(T).$
$N(T)$ denotes the kernel of $T$.
We equip the quotient space $X/N(T)$ with the quotient norm  $|| \cdot ||$. Then this space is a Banach space.
Further we define the linear mapping $T_0: X/N(T) \to Y$ by $T_0(x+N(T)) :=Tx.$
Show that $T_0$ is bounded and bijective. Since $X/N(T)$ and $Y$ are Banach spaces, $T_0^{-1}$ is bounded.
If $y \in Y$ there exists $x_0 \in X$ with $Tx_0=y$, hence $y=T_0(x_0+N(T))$, thus
$$x_0+N(T)=T_0^{-1}y.$$
Therefore
$$||x_0+N(T)|| \le ||T_0^{-1}|| \cdot ||y||_Y.$$
The definition of the quotient norm now gives some $x \in x_0+N(T)$ such that
$$||x||_X \le (||T_0^{-1}|| +1)\cdot ||y||_Y.$$
Since $x-x_0 \in N(T)$, we have $Tx=Tx_0=y.$
