Decomposition $Y=\text{im}T \oplus N$ for Fredholm operator $T$

Asked Jul 29 '19 at 19:54

Active Jul 29 '19 at 19:54

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I don't get how, from the definition of Fredholm Operator below, we get $Y=\text{im}T \oplus N$ with $N$ closed?

Example 11.6(1) states Finite-Dimensional subspaces are always complemented.

Example 11.6(2) states Finite-Codimensional Closed subspaces are complemented.

Example 11.4(1) states if $T:X\rightarrow Y$ is an operator, $X,Y$ are Banach, and $Y=\text{im}T \oplus N$ with $N$ closed, then $\text{im}T$ is also closed.

How can 11.4(1) help us? it seems to assume precisely what we want to prove...

asked Jul 29 '19 at 19:54

An old man in the sea.

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finite dimensional spaces are closed. idk how those examples are relevant. – mathworker21 Jul 29 '19 at 20:16
@mathworker21 $imT$ may not be finite-dimensional, from what I understood, but has finite codimension. – An old man in the sea. Jul 29 '19 at 20:38
maybe the following argument works: $Y/imT \cong ker T$, so since $ker T$ closed, $Y/imT$ closed (the isomorphism should be continuous), so $im T$ closed – mathworker21 Jul 29 '19 at 20:43
@mathworker21 the only theorem similar to that, which I know is the 1st isomorphism theorem, but with $kerT$ interchanged with $imT$... Is that a known theorem or property? – An old man in the sea. Jul 29 '19 at 21:02
oops, of course. I'll come back to this. – mathworker21 Jul 29 '19 at 22:38
@mathworker21 by the way, check this my other question related to this... I think it can help us both. ;) https://math.stackexchange.com/questions/3307828/why-is-the-closed-im-t-in-the-definition-of-fredholm-operator-redundant – An old man in the sea. Jul 29 '19 at 23:08

Decomposition $Y=\text{im}T \oplus N$ for Fredholm operator $T$

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