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A projection with finite-dimensional kernel is Fredholm

Example

Assume the Axiom of Choice (The Axiom of Choice). Let $N$ and $Y$ be real Banach spaces with $\dim N < \infty$ (Banach space), and let $X = N \oplus Y$ be their topological direct sum with bounded projections (A complemented closed subspace of a normed space), where the direct sum is second countable (Second countability: an at most countable basis for the topology) — for instance this holds whenever $N$ and $Y$ are second countable (Assuming countable choice, a countable product of second countable spaces is second countable). Then $X$ and $Y$ are $C^{\infty}$ Banach manifolds in the sense of Countable base Banach manifold and smooth map, and the projection $p : X \to Y$ onto $Y$ along $N$ is a smooth Fredholm map (Fredholm map between Banach manifolds) of index $\dim N$ , and its local finite-dimensional reduction (Local finite-dimensional reduction for a Fredholm map) has zero obstruction space: in suitable coordinates it is the projection $(u, v) \mapsto (u, 0)$ onto the range factor of the splitting $ran p \oplus C$ , with the complement coordinate set to zero.

Facts & Assumptions

Given: Real Banach spaces $N, Y$ with $\dim N < \infty$ and second countable topological direct sum $X = N \oplus Y$ with bounded projections, and the projection $p : X \to Y$ onto the second factor.

[L1]

In a topological direct sum $X = N \oplus Y$ every $x$ decomposes uniquely as $x = n + y$ and the coordinates $n = q (x)$ , $y = p (x)$ are bounded linear; here $\ker p = N$ is finite dimensional by hypothesis and $ran p = Y$ (A complemented closed subspace of a normed space).

[L2]

A bounded linear map is differentiable everywhere with derivative itself, and is smooth of class $C^{\infty}$ as a map of Banach manifolds (Fréchet derivative between Banach spaces, C k map between Banach spaces).

[L3]

Fredholm operator and index: finite-dimensional kernel, closed range and finite-dimensional cokernel, index $\dim \ker - \dim coker$ (Fredholm operator cokernel and index); the local finite-dimensional reduction produces coordinates in which a Fredholm map is $(u, v) \mapsto (u, g (u, v))$ with $u$ ranging over an open subset of the range, $v$ over an open subset of the finite-dimensional kernel, and $g$ taking values in a finite-dimensional complement of the range (Local finite-dimensional reduction for a Fredholm map).

Verification

technique · direct

1.1L1L3algebra

By [L1] the map $p$ is bounded linear with $\ker p = N$ finite dimensional, $ran p = Y$ closed, and $coker p = Y / Y = {0}$ finite dimensional; hence $p$ is Fredholm at every point with index $\dim N - 0 = \dim N$ by [L3].

2.1step 1.1L2

The map $p$ is smooth and $D p (x) = p$ for every $x$ by [L2], so $p$ is a smooth Fredholm map of index $\dim N$ by [step 1.1].

2.2step 1.1L1L3

For the reduction, take the splitting $X = \ker p \oplus Y$ and the range complement $C = {0}$ ; the normal form of [L3] reads $(u, v) \mapsto (u, g (u, v))$ with $g$ valued in the zero space, so $g \equiv 0$ and the obstruction space is trivial; the coordinates are those of the direct sum itself, and no nontrivial correction term is produced.

3.1step 2.1step 2.2∎

Thus $p$ is a smooth Fredholm map of index $\dim N$ whose local reduction has zero obstruction space, as claimed.

Depends on

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Sources

Alberto Abbondandolo and Pietro Majer, Lectures on the Morse Complex — §2.11 (standard reference, not scraped)