A subset U ⊆ X/~ is declared open in the quotient topology when which condition holds?
Aq(V) = U for some open set V ⊆ X
Bq⁻¹(U) is open in X
CU is an open ball of equivalence classes under some metric
DU is contained in the image of an open set of X
The quotient topology is defined by declaring U ⊆ X/~ open if and only if its PREIMAGE q⁻¹(U) is open in X. This is exactly what forces the quotient map q to be continuous. Option A is the most tempting wrong answer: it uses images rather than preimages, but q is not generally an open map — the image of an open set need not be open in the quotient. Preimages are the right tool because continuity is defined via preimages.
Question 2 Multiple Choice
A student claims: 'The quotient topology is the coarsest topology on X/~ that makes q continuous.' What is wrong with this claim?
ANothing — this is a correct characterization of the quotient topology
BIt is the finest (largest) topology making q continuous, not the coarsest
CThe quotient map is never continuous, so no such topology exists
DCoarseness and fineness are not defined for quotient spaces
The quotient topology is the FINEST (largest) topology on X/~ making q continuous — it declares as many sets open as possible while still maintaining continuity of q. A coarser topology (fewer open sets) would also make q continuous, since fewer open sets means fewer preimage conditions to check. The indiscrete topology is the coarsest that makes q continuous, but it is trivial and loses all structure. The quotient topology is the natural maximum — the richest topology consistent with continuity of q.
Question 3 True / False
In the quotient topology, the open sets of X/~ are exactly the images q(U) of open sets U in X.
TTrue
FFalse
Answer: False
This reverses the direction: open sets in X/~ are determined by PREIMAGES, not images. A set V ⊆ X/~ is open iff q⁻¹(V) is open in X. The map q is not generally an open map — it does not need to send open sets to open sets. A concrete example: when [0,1] is collapsed by identifying 0 ~ 1 to form S¹, there are open sets in [0,1] whose images under q are not open in the quotient.
Question 4 True / False
When building the torus by identifying opposite edges of the unit square, the quotient topology is well-defined even though the resulting space is not a subset of any Euclidean space used in the construction.
TTrue
FFalse
Answer: True
The quotient topology is defined purely in terms of the equivalence relation and the topology on the original space — it does not require the quotient space to be embedded in any ambient Euclidean space. The torus, projective plane, Klein bottle, and many other spaces are naturally constructed as quotients, and their topology is fully determined by the quotient construction. This is precisely the power of the framework: it defines a topology on X/~ intrinsically, not by reference to an embedding.
Question 5 Short Answer
Why is the quotient topology defined using preimages of the quotient map q rather than images?
Think about your answer, then reveal below.
Model answer: Because continuity of q requires that preimages of open sets are open. Defining open sets in X/~ via preimages of q is exactly what forces q to be continuous — and the quotient topology is the finest topology with this property. Images of open sets under q need not be open because q is not generally an open map.
This connects to the fundamental definition of continuity in topology: f is continuous iff the preimage of every open set is open. By defining the quotient topology so that U ⊆ X/~ is open iff q⁻¹(U) is open, we are effectively building continuity of q into the definition of open sets. This also gives the quotient topology its universal property: a function f: X/~ → Y is continuous iff f ∘ q: X → Y is continuous, which is what makes the quotient the 'right' domain for maps that respect the equivalence relation.