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Is this true?

A continuous bijection is a homeomorphism

No — it is false.

The counterexample

id : $(R$ , discrete $) \to (R$ , usual), or t ↦ (cos t, sin t) from $[0, 2 π)$ to $S^{1}$

Needs compact domain and Hausdorff codomain.

The kind of mistake this is

Property not inherited

A property assumed to pass to subobjects, quotients, or through a chain. It does not.

Drill statements like this

Others that fail the same way

From Topology › Continuity, homeomorphism, separation axioms

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