Skip to main content
added 83 characters in body
Source Link

Let $E$ be the union of the coordinate axes in $\mathbb R^2$, and let $\xi:E\to\mathbb R$ be the projection on the first coordinate. Then $\Gamma(\xi)$ is a trivial sheaf, so it is isomorphic to $C(\mathbb R,F)$ with $F$ any one-point space. Yet $\xi$ is not locally trivial. (This can be modified so that the fiber at $0$ of $\xi$ is pretty much anything...)

If this is too trivial an example, let $F$ be any manifold, let $E$ be as before, and let $\xi=E\times F\to\mathbb R$ such that $\xi((x,y),f)=x$. Then $\Gamma(\xi)\cong C(\mathbb R,F)$, just as before.

Let $E$ be the union of the coordinate axes in $\mathbb R^2$, and let $\xi:E\to\mathbb R$ be the projection on the first coordinate. Then $\Gamma(\xi)$ is a trivial sheaf, so it is isomorphic to $C(\mathbb R,F)$ with $F$ any one-point space. Yet $\xi$ is not locally trivial.

If this is too trivial an example, let $F$ be any manifold, let $E$ be as before, and let $\xi=E\times F\to\mathbb R$ such that $\xi((x,y),f)=x$. Then $\Gamma(\xi)\cong C(\mathbb R,F)$, just as before.

Let $E$ be the union of the coordinate axes in $\mathbb R^2$, and let $\xi:E\to\mathbb R$ be the projection on the first coordinate. Then $\Gamma(\xi)$ is a trivial sheaf, so it is isomorphic to $C(\mathbb R,F)$ with $F$ any one-point space. Yet $\xi$ is not locally trivial. (This can be modified so that the fiber at $0$ of $\xi$ is pretty much anything...)

If this is too trivial an example, let $F$ be any manifold, let $E$ be as before, and let $\xi=E\times F\to\mathbb R$ such that $\xi((x,y),f)=x$. Then $\Gamma(\xi)\cong C(\mathbb R,F)$, just as before.

Source Link

Let $E$ be the union of the coordinate axes in $\mathbb R^2$, and let $\xi:E\to\mathbb R$ be the projection on the first coordinate. Then $\Gamma(\xi)$ is a trivial sheaf, so it is isomorphic to $C(\mathbb R,F)$ with $F$ any one-point space. Yet $\xi$ is not locally trivial.

If this is too trivial an example, let $F$ be any manifold, let $E$ be as before, and let $\xi=E\times F\to\mathbb R$ such that $\xi((x,y),f)=x$. Then $\Gamma(\xi)\cong C(\mathbb R,F)$, just as before.