Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In R.O.Wells book "Differential Analysis on Complex Manifolds" p. 44 proof of Theorem 2.2 part b) the author claims that any two sections of an etale space which agree at a point agree in some neighborhood of that point (etale space is a (possibly non-Hausdorff) $Y$ with the surjection $p\colon Y \rightarrow X$ which is a local homeomorphism). Why is that true?

share|improve this question
2  
Why is this tagged "soft-question"? –  Charles Staats May 5 '10 at 3:51
    
I thought it should be trivial, since the author assumes it so casually... –  noname May 5 '10 at 8:39
    
The soft-question tag is used more on this site for "metamathematical" questions (not sure if that's the right word). I've removed it. –  j.c. May 5 '10 at 17:18
add comment

1 Answer 1

up vote 6 down vote accepted

First note that if $p_1:Y_1 \to X$ and $p_2:Y_2 \to X$ are two etale spaces over $X$ then any morphism of etale spaces $f:Y_1 \to Y_2$ (where by morphism of etale spaces I mean that $f$ is continuous and that $p_2\circ f = p_1$) makes $Y_1$ an etale space over $Y_2$.

(Proof: Let $y \in Y_1$, and let $V$ be a n.h. of $f(y)$ in $Y_2$ such that $p_2:V \to p_2(V)$ is a homeo. Note that a local homeo. is open by definition, and so $p_2(V)$ is also open in $X$. Now $f^{-1}(V)$ is an open subset of $Y_1$ containing $y$, and so we may find a n.h. $U$ of $y$ in $Y_1$, contained in $f^{-1}(V)$, such that $p_1: U \to p_1(U)$ is a homeo. Since $p_1 = p_2\circ f$, and since $f(U) \subset V$, we find that $f: U \to f(U)$ is a homeo. and that $f(U)$ is open in $Y_2$. Thus $f$ is a local homeo., as required.)

Now suppose that $f_1,f_2:Y_1 \rightarrow Y_2$ are two morphisms of etale spaces, and that $f_1(y) = f_2(y)$ for some $y \in Y_1$. Then $f_1$ and $f_2$ coincide in some n.h. of $y$.

(Proof: Choose some sufficiently small n.h. $V$ of $f_1(y) = f_2(y)$ in $Y_2$ such that $p_2:V \to p_2(V)$ is a homeo. Then choose a n.h. $U$ of $y$ contained in $f_1^{-1}(V) \cap f_2^{-1}(V)$ such that $f_i:U \to f_i(U)$ is a homeo. for both values of $i$. Since each $f_i(U)$ is contained in $V$, we see that $p_2: f_i(U) \to p_2\circ f_i(U) = p_1(U)$ is a homeo. as well, and hence that $p_1 = p_2\circ f_i: U \to p_1(U)$ is a homeo.

We then find that, on $U$, the map $f_i$ (for either value of $i$) can be computed as the composite of $p_1:U \to p_1(U)$ and $p_2^{-1}: p_1(U) \to f_i(U) \subset V$. Thus the two maps $f_i$ coincide on $U$.)

Now if $p:Y \to X$ is an etale space and $\sigma$ is a section, we can think of $\sigma$ as a morphism from the etale space $X \to X$ (i.e. $X$ thought of as the trivial etale space over itself) to the etale space $Y \to X$. The preceding result then shows that if two sections coincide at a point, they coincide in a n.h. of that point.

share|improve this answer
    
Very nice. Thank you! –  noname May 5 '10 at 10:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.