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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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    For those who want to take a look at it: Page 27 [here.](http://books.google.com/books?id=fJzlLkHBYTYC&printsec=frontcover&dq=Max+Karoubi+%22K-Theory%22&source=bl&ots=s8I85R68xD&sig=NrkH5wPQqAnzcvk8awbwyDvQJMc&hl=de&ei=hldUTZHkHMPpOfOZ-KIF&sa=X&oi=book_result&ct=result&resnum=2&ved=0CCUQ6AEwAQ#v=snippet&q=px%20p0%20%22is%20the%20trivial%20bundle%22%20%22Since%20the%20question%22&f=false)2011-02-10

1 Answers 1

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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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    You$r$ alternate solution does the job. I just have to re-work his proof a bit. Thanks.2011-02-22