False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-31

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The intrinsic topology of an immersed submanifold need not be the subspace topology

Statement

False claim: an immersed submanifold always carries the subspace topology of its image in the ambient manifold.

Facts & Assumptions

Given: The same componentwise inclusion $F : M \to R^{2}$ from the countable family of concentric circles used above.

[F1]

An immersed submanifold is a manifold with an injective immersion into the ambient manifold; its intrinsic topology is not defined to be the subspace topology (Immersed submanifolds).

[F2]

Embedded submanifolds, by contrast, do use the subspace topology (Embedded submanifolds and slice charts).

[L1]

The disjoint-union source is a smooth manifold and each circle is smooth (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

1.1F1L1given

By [L1], the source $M$ is a smooth one-manifold, and the map $F$ is an injective immersion componentwise. Hence $(M, F)$ is an immersed submanifold in the sense of [F1].

2.1F2step 1.1

In the intrinsic topology of $M$ , each circle component is open because $M$ is a disjoint union. In the subspace topology on $F (M) \subseteq R^{2}$ , the unit circle component is not open because every neighbourhood of one of its points meets infinitely many outer circles. Thus the two topologies differ.

3.1F1step 1.1step 2.1∎

Therefore an immersed submanifold need not carry the subspace topology of its image.

Depends on

Used by

Countably many concentric circles give an injective immersion that is not an embedding Counterexample
The subspace topology on a leaf is always its manifold topology False statement

Dependency tree · two levels

34 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, Immersed Submanifolds (standard reference, not scraped)