False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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An injective immersion need not be an embedding

Statement

False claim: every injective immersion is an embedding.

Facts & Assumptions

Given: The disjoint union $M$ of one unit circle and the circles of radii $1 + 1 / n$ for $n \geq 1$ , together with the componentwise inclusion $F : M \to R^{2}$ .

[F1]

A smooth embedding is an injective immersion that is a homeomorphism onto its image with the subspace topology (Smooth embeddings).

[F2]

An immersed submanifold only requires an injective immersion; its intrinsic topology need not be the subspace topology (Immersed submanifolds).

[L1]

Countable disjoint unions of fixed-dimensional smooth manifolds are smooth manifolds, and each circle is a smooth embedded one-manifold (Countable disjoint unions of fixed-dimensional smooth manifolds are smooth manifolds, A regular level set is an embedded submanifold, For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

Refutation

technique · direct

1.1

By [L1], $M$ is a smooth one-manifold, and on each circle component the map $F$ is the usual inclusion into $R^{2}$ , hence an immersion. The images of different components are distinct circles, so $F$ is injective.

L1given

2.1

The image $F (M)$ is the union of the concentric circles. Every Euclidean neighbourhood of a point on the unit circle meets infinitely many outer circles, so the unit-circle component is not open in the subspace topology of $F (M)$ . But that component is open in the disjoint-union topology of $M$ . Therefore $F : M \to F (M)$ is not a homeomorphism.

F2step 1.1

3.1

Step 1.1 gives an injective immersion, while step 2.1 shows that the homeomorphism clause in [F1] fails. Hence $F$ is not an embedding, which refutes the claim.

F1step 1.1step 2.1∎

Depends on

Used by

Countably many concentric circles give an injective immersion that is not an embedding Counterexample

Dependency tree · two levels

36 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John M. Lee, Introduction to Smooth Manifolds, Embeddings (standard reference, not scraped)