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Show that a continuous injective map f: (0, 1) to R is an embedding.

Show that a continuous injective map f: (0, 1) to R is an embedding.
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Answer #1

Let f : X → Y be an injective continuous function between topological spaces. Then f is also a continuous bijective function f : X → f(X), where you replace the codomain with the range.

We say that a continuous injective function f : X → Y is an embedding if it the bijective continuous function f : X → f(X) has continuous inverse.

The map that sends R to the y-axis in is not a homeomorphism, but it is a homeomorphism onto the range.

The map f : [0, 1) → R which sends x to is continuous, as it is composed out of algebraic and transcendental functions. It’s also injective, as you can check. The range of f is the unit circle T = {x ∈ R : ||x|| = 1} in R . The inverse map : T → [0, 1) is NOT continuous, as the pre-image of the open set [0, 1/2) is not open in the circle (the circle inherits the Euclidean topology from R)

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