Are injective operators invertible?

Question

I am little worried about the notion of invertible operators.

Let $E,F$ be two Banach spaces and $A:D(A)\subset E \to F$ be an operator. We have the following lemma:

  If A is closed and injective, then its inverse is closed.

We say that $A$ is invertible if and only if $A$ is bijective map. i.e there exists an operator $B:F \to D(A)$ such that $BAu=u$ for every $u \in D(A)$ and $ABv=v$ for every $v \in F$.

In the lemma they are not mentioning surjectivity!

Answer 1

When $A : D(A) \to F$ is injective, it's bijective on the range $R(A)$, and then you can define its inverse $A^{-1}: R(A) \to D(A)$.