in the Higson and Roe's book "analytic K-homology" just after the definition of the Fredholm operator there is a remark (2.1.3 you can see at it onlin at Google books (click here)) which claims that in the definition of Fredholm operators the condition of being closed for image of operator is superfluous. It seems that in the proof $cokernel(T)$ is supposed to be $\frac{H_2}{Image(T)}$ instead of $\frac{H_2}{\overline{Image(T)}}$ which cause the newly defined operator $\tilde{T}$ to be immediately surjective. (which is not obviouse for me when I take $coker(T)= \frac{H_2}{\overline{Image(T)}}$ (as it is expected when we consider coker at the category of Hilbert spaces with bounded operators)).

I was wondering if there is anybody help me to find out what is goning on in this proof or prove or disprove the statement in a clear way?

Edit:

So You agree with me that this proof is true just if we take $cokernel_1(T)=\frac{H_2}{Image(T)}$.

Now here is the question: what if we put $cokernel_2(T)=\frac{H_2}{\overline{Image(T)}}$? I mean if $cokernel_2(T)$ is finite dimension can we say that $cokernel_1(T)$ is finite dimensional also?

In other word: Is there any bounded operator $T:H_1\to H_2$ which has finite dimensional $coker_2(T)$ (please note the index 2) and $ker(T)$ but it is not a Fredholm operator?