TITLE: Why is the space of surjective operators open?
QUESTION [12 upvotes]: Suppose $E$ and $F$ are given Banach spaces. Let $A$ be a continuous surjective map. Why is there a small ball around $A$ in the operator topology, such that all elements in this ball are surjective?

REPLY [16 votes]: A bounded linear operator $A: E \to F$, where $E$ and $F$ are Banach spaces, is surjective if and only if there is $c > 0$ such that for all $\phi \in F^*$,
$\|A^* \phi\| \ge c \|\phi\|$ (see e.g. Rudin, "Functional Analysis", Theorem 4.15).
Now if $A$ is such an operator, so is $B$ for $\|A-B\| < c$, since 
 $$\|B^* \phi\| \ge \|A^* \phi\| - \|A^* - B^*\| \|\phi\| \ge (c - \|A-B\|) \|\phi\|$$