Is the "closedness of the image of operator" needed in the defintion of Fredholm operators?

Question

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?

Answer 1

No, the proof is correct, and it's taking $\mathrm{coker}\ T = H_2/\mathrm{im}\ T$. The essence of the argument is that $V\oplus\mathbb{C}^n$ contains $V$ as a closed subspace, regardless of the norm on $V$, and that $H_2$ is "close enough" to $(\mathrm{im}\ T)\oplus\mathbb{C}^n$.

The full argument goes like this: Let $T: H_1\to H_2$ be Fredholm (the $H_i$ are Hilbert spaces). WLOG $T$ is injective, otherwise replace $H_1$ with $(\mathrm{ker}\ T)^\perp$. Choose $v_1,\dots, v_n\in H_2$ which represent a basis of $\mathrm{coker}\ T$, and use them to define a linear function $\mathbb{C}^n\to H_2$. Combine this with $T$ to get a linear map $\overline{T}: H_1\oplus\mathbb{C}^n\to H_2$.

Now $\overline{T}$ is a linear isomorphism, so its inverse $\widetilde{T}$ is also a linear isomorphism. $\overline{T}$ is also continuous, so the graph of $\overline T$ is a closed subset of $H_1\oplus\mathbb{C}^n\oplus H_2$, so the graph of $\widetilde T$ is a closed subset of $H_2\oplus H_1\oplus\mathbb{C}^n$, which makes $\widetilde T$ continuous by the Closed Graph Theorem. (This last is not an obvious statement; it's a consequence of the Baire Category Theorem). Finally, $\mathrm{im}\ T$ is precisely the pullback of $H_1$ through $\widetilde T$, making it closed.

As to why most sources don't present it this way… I think in most cases where somebody wants to prove that a given operator $T$ is Fredholm, it's just easier to establish that $\mathrm{coker}\ T$ is finite-dimensional after proving that $\mathrm{im}\ T$ is closed.

Answer 2

If $X$ and $Y$ are Banach spaces, $T:D(T)\subset X\to Y$ is a closed operator and there exists a closed subspace $N$ of $Y$ such that $\operatorname{Im}(T)\cap N=\{0\}$ and $\operatorname{Im}(T)\oplus N$ is closed, then $\operatorname{Im}(T)$ is closed.

See Theorem IV.5.10 in A.E. Taylor and D.C. Lay. Introduction to functional analysis, 2nd ed. Krieger 1980.

Answer 3

If the co-dimension is finite in the "algebraic" sense, rather than topological, this is indeed true, see the little beautiful book by S. Krein "Linear equations in Banach spaces" (somewhere in the first dozen pages, I think), as well as remark on p. 125 in "Banach bundles and linear operators" by M. Zaidenberg et al, Russian Math. Surveys 30:5 (1975), 115-175