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I am reading Max Karoubi's "K-Theory" and I think I'm overlooking some trivial fact. We have a vector bundle $E\rightarrow X$ and a morphism $p:E\rightarrow E$ with $p^2=p$. He is showing that $\ker p$ is locally trivial. First he assumes that $E=X\times V$ for a vector space $V$. Here's where I'm stuck:

He defines $f:X\longrightarrow \operatorname{End}(V)$ by

$$ f(x)=1-p_{x_0}-p_x+2p_xp_{x_0}, $$

where $p_x$ is the restriction of $p$ to the fiber over $x$ and $x_0$ is a basepoint. The claim is that $p_{x_0}\circ f(x)=f(x)\circ p_x$. When I compute both sides I get

$$ 2p_{x_0}p_xp_{x_0}-p_{x_0}p_x=2p_{x}p_{x_0}p_{x}-p_{x_0}p_x $$

which says

$$ 2p_{x_0}p_xp_{x_0}=2p_{x}p_{x_0}p_{x}. $$

Why is that true? Thanks.

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Maybe I'm being really dumb, but doesn't $p_x: V\to V$? What exactly is meant by equality of maps in this sense? It seems $p_{x_0}\circ f: X\to End(V)$, but $f\circ p_x: V\to $ something, I'm not sure. Sorry if I appear to have no idea what I'm talking about. –  Matt Feb 10 '11 at 17:47
    
@Matt: You are not being foolish. It should read $$p_{x_0}\circ f(x)=f(x)\circ p_x.$$ It has been corrected. Thanks. –  Joe Johnson 126 Feb 10 '11 at 18:29
    
For those who want to take a look at it: Page 27 here. –  Rasmus Feb 10 '11 at 21:29

1 Answer 1

up vote 2 down vote accepted

That's probably a typo. Try $p_x f(x)= f(x) p_{x_0}$ or, alternatively, $f(x)=1-p_{x_0}-p_x +2 p_{x_0} p_{x}$ in the rest of the proof (which I can't see).

As given, the equality is not true. Think of 2 lines, say of slope 2 and 1/2, in R^2 and the projections along vectors (0,1) and (1,0). Then the image of the first composition is all of the first line, and of the second one all of the second line. No way such maps can be equal.

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Your alternate solution does the job. I just have to re-work his proof a bit. Thanks. –  Joe Johnson 126 Feb 22 '11 at 15:01

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