Manuscript XIX-D@1 · XIX-D.tex · proposition · 1f47e33b

Proposition: Projection leaves a precise unresolved component

Counterterm images, finite cancellation, and projection › Projected identities and projected fixed points · lines 812–832 · proof included or linked

XIX-D@1#prop:projection

Let $P : E \to E$ satisfy $P^{2} = P$ . For $d \in E$ , the equality $P d = 0$ implies only $d \in ker P$ . A projected fixed-point equation

P F (x) = P x

implies only $F (x) - x \in ker P$ , not $F (x) = x$ .

Proof. Both assertions follow from the kernel definition. For a nonzero example, let

E = C^{2}, P (x, y) = (x, 0), F (x, y) = (x, y + 1) .

Every point satisfies the projected equation, but none is a fixed point of $F$ .

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