# Why Are These Two Morphisms the Same?

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

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).