Continuity through inverse images

Try

A constant map from the real line to itself is continuous, but it sends the open interval (0,1) to a singleton, which is not open. This small example exposes a frequent reversal: continuity concerns inverse images of open sets, not images.

If the output is required to remain within a neighborhood V of f(x), continuity asks for an input neighborhood U of x whose entire image lies in V. It is a containment guarantee near each point, not a statement about the shape of every image set.

Predict

Predict whether changing the domain topology to a finer one makes continuity easier or harder for a fixed map into the same target. It makes continuity easier: there are more possible open inverse images. Making the target topology finer introduces more sets whose preimages must be checked, so that can make continuity harder.

Use the identity map between the usual and discrete real line to test both directions.

Experiment

Check maps using inverse images.

Map	Reason for continuity or failure
Any map from a discrete space	Every inverse image is open
Any map into an indiscrete space	Only empty and whole inverse images are required
Usual R to discrete R, identity	A singleton target-open has nonopen preimage
Discrete R to usual R, identity	Every required preimage is open

The same underlying identity function has different continuity behavior when its source and target topologies are exchanged.

Understand

A map f:X→Y is continuous when f⁻¹(V) is open in X for every open V in Y. The pointwise neighborhood condition is equivalent. If preimages are open, f⁻¹(V) itself is an input neighborhood around each point mapping into V. Conversely, if every point of a preimage has an open neighborhood contained in that preimage, the preimage is their union and hence open.

Taking complements proves the closed-set criterion: f is continuous exactly when inverse images of closed sets are closed. Composition is continuous because (g∘f)⁻¹(W)=f⁻¹(g⁻¹(W)).

There is also a closure criterion: f is continuous exactly when f(closure A)⊆closure(f(A)) for every A⊆X. In the forward direction, pull any neighborhood of f(x) back to a neighborhood of x and use x∈closure A. For the reverse, take a closed C⊆Y and A=f⁻¹(C). The assumed containment gives f(closure A)⊆closure(f(A))⊆C, so closure A⊆A. Hence A is closed and the closed-set criterion applies.

A useful pasting lemma says that continuous maps on a finite closed cover, agreeing on overlaps, define a continuous whole-space map. For any target closed set, its preimage is a finite union of sets closed in the covering pieces; because those pieces are closed in X, each is closed in X. Finiteness makes the union closed.

Apply

Use pasting to prove the absolute-value function is continuous by splitting R into its two closed half-lines and using x and −x on the appropriate pieces. Verify agreement at zero before invoking the lemma.

Diagnose “An arbitrary union of continuous pieces is continuous.” The pieces need a valid gluing hypothesis. A finite closed cover works, as does an arbitrary open cover with the corresponding open-preimage proof. An arbitrary closed cover does not support the finite-union argument.

Continue

Homeomorphisms will next require continuity in both directions, not merely a continuous bijection. Optional depth: prove the open-cover pasting lemma directly, and compare why open unions allow arbitrarily many pieces while the simple closed-cover proof needs finitely many.