Questions: Quotient Maps and Identification Spaces

Question 1

You have a surjective continuous map f from [0,1] (compact) to a Hausdorff space Y. Without checking preimages of every open set, can you conclude f is a quotient map?

• A. No — compactness and Hausdorff are irrelevant; you must verify the open-set condition directly
• B. Yes — any surjective continuous map is automatically a quotient map
• C. Yes — a surjective continuous map from a compact space to a Hausdorff space is automatically a quotient map
• D. Only if f is also injective, making it a homeomorphism

Answer: (C)

The key theorem states: any surjective continuous map from a compact space to a Hausdorff space is a quotient map. The proof uses the fact that in this setting, closed sets in X map to closed sets in Y (a continuous map from compact to Hausdorff is a closed map), which implies the quotient topology condition. This theorem is enormously useful in practice — it lets you verify quotient maps without checking every open set, as long as you can establish compactness and Hausdorff separation.

Question 2

Both endpoints of [0,1] are identified: 0 ~ 1, while all other points are equivalent only to themselves. What is the resulting quotient space?

• A. A closed interval [0,1] — identifying two points does not change the topology significantly
• B. A circle — identifying the endpoints bends the interval and glues them into a single point
• C. Two separate components — the identified endpoints form one piece, the interior another
• D. An open interval (0,1) — the identified point at the boundary is removed from the space

Answer: (B)

Identifying the endpoints of [0,1] produces a space homeomorphic to S¹ (the circle). Intuitively, the interval is 'bent' so its two ends meet at a single point, forming a loop. The quotient topology makes this rigorous: an open set in the quotient must have a preimage that is open in [0,1], and the neighborhoods of the identified endpoint must include open sets from both ends of the interval simultaneously — exactly the behavior of a neighborhood of a point on a circle.

Question 3

Most surjective continuous map from one topological space to another is a quotient map.

• T. True
• F. False

Answer: False

A continuous bijection that is not a homeomorphism provides a counterexample: the map f: [0,1) → S¹ given by f(t) = e^{2πit} is a continuous bijection, but its inverse is not continuous (the circle topology is 'finer' in the relevant sense). So f is surjective and continuous but not a quotient map — there are open sets in S¹ whose preimages are not open in [0,1). A quotient map requires the stronger condition: V is open in Y if AND ONLY IF f⁻¹(V) is open in X.

Question 4

If q: X → Y is a quotient map and X is compact, then Y is compact.

• T. True
• F. False

Answer: True

Continuous images of compact spaces are compact — this is a standard theorem in topology. Since a quotient map is in particular a continuous surjection, Y = q(X) is a continuous image of the compact space X, hence compact. This is one of the few 'nice' properties that quotient maps reliably preserve; Hausdorff separation, for instance, can be lost under quotient maps.

Question 5

What property defines a quotient map, and why does this definition give the codomain 'exactly the right topology'?

Model answer: A quotient map q: X → Y is a surjection where V ⊆ Y is open if and only if q⁻¹(V) is open in X. This gives Y the finest topology making q continuous: if you made the topology coarser (fewer open sets), you would lose open sets whose preimages are open; if you made it finer (more open sets), some open set in Y would have a non-open preimage, making q discontinuous. The 'exactly right topology' captures precisely the open structure of X that is visible through the identification — open sets in Y correspond exactly to saturated open sets in X.

The 'if and only if' is crucial. The 'if' direction (open preimage implies open in Y) says the topology is fine enough to include all opens arising from X. The 'only if' direction (open in Y implies open preimage) says the topology is not finer than necessary — every open set in Y must be 'witnessed' by an open set in X. Together, these conditions uniquely characterize the quotient topology as the finest topology making q continuous.