Topology paper 2022

1. 1. Define the operation of connected sum of surfaces.

      The connected sum of and is the surface where and are embedded discs, and the equivalence relation ∼ identifies and via some homeomorphism [Not required: If and are surfaces with boundary, then and should not meet the boundary].

   2. State the classification theorem of surfaces in terms of connected sums of known surfaces.

      Every closed surface is homeomorphic to either , a connected sum of copies of the torus , or a connected sum of copies of the projective plane . Moreover, the surfaces in the above list are pairwise non-homeomorphic.

2. Prove that these two spaces are homeomorphic: (the connected sum of two Möbius bands, and the Klein bottle with two discs removed). [Hint: describe (and prove) how the Möbius band is related to , then describe (and prove) how two copies of are related to the Klein bottle.]

   The Möbius band is homeomorphic to with one disc removed: Lemma 5.12
   To see this, consider the following construction of : and remove this disc:
   So the connected sum of two Möbius bands is homeomorphic to with two discs removed.
   It remains to show that is homeomorphic to a Klein bottle: Example 5.22 The boundary words xy⁻¹xy and xxyy represent homeomorphic surfaces.

3. Let be the unit circle.
   1. Is ℝ homeomorphic to a subspace of ? If yes:
      α) is it homeomorphic to an open subspace of ?

      Yes: minus a point is homeomorphic to an open interval. The latter is homeomorphic to ℝ.

      β) is it homeomorphic to a closed subspace of ?

      No: Any closed subset of is compact, but ℝ is not compact.

   2. Is homeomorphic to a subspace of ℝ ?
      If yes: α) is it homeomorphic to an open subspace of ℝ ? β) is it homeomorphic to a closed subspace of ℝ ?

      No: Any subset of ℝ (with at least three points) has the property that it becomes disconnected after removing some point. But is connected and remains connected after removing any point.

4. 1. Prove that is open in ℝ if and only if is open in for all .

      If is open, then is open in by definition of the subspace topology. Assume that is open in for all . We need to show that for every point there exists such that . Pick any interval containing in its interior. Since is open in , there exists such that . In particular, .

   2. Let be a topological space. Is it true that a subset is open in if and only if is open in for all closed subsets ?
      α) Prove that the answer is "yes" when is Hausdorff.

      (If is open, then is open in by definition of the subspace topology.) Assume is open in for all closed subsets . We'll show that is open by proving that every point admits an open neighbourhood contained in . Pick distinct from (if has only one point, then is trivially open). Since is Hausdorff, there exist disjoint open subsets with and . Let , so thatBy assumption, is open in . Therefore is open in . Since is open in is also open in . is the desired neighbourhood of .

      β) Prove that the answer is "no" in general.

      Let be the two point space equipped with the indiscrete topology, and let be a subspace consisting of just one of the two points. is not open, but is open in (and is the only closed subset of which is not the whole space).

1. 1. Define what it means for a topological space to be compact. [If you use words such as "cover" and "subcover", you must also define these words.]

      A topological space is compact if every open cover (an open cover is a collection of open sets whose union is ) admits a subcover (a subcover is a sub-collection whose union is still ) where is finite.

   2. A topological space is called locally compact if for every point there exists open and compact such that .
      α) Prove that ℝ with its usual topology is locally compact.

      Given , we have . In other words, is a compact neighbourhood of .

      β) Is ℚ (equipped with the subspace topology from ℝ) locally compact? Justify.

      No point admits a compact neighbourhood. If is a neighbourhood of , then for some . Since is not complete, it is not compact, but is closed in , so is not compact.

2. Let be the closed unit disc.
   1. Prove that any bounded convex polygon is homeomorphic to .

      We may assume without loss of generality (by performing a translation) that contains 0 in its interior. Let be defined by the requirement that . Note that this is well-defined: since is convex, the intersection is also convex, in particular connected. The function is clearly continuous (it is piecewise algebraic). The map given, in polar coordinates, by is the desired homeomorphism. It is continuous at non-zero points because is continuous. It is continuous at zero because is bounded. Finally, it maps bijectively onto , and is thus a homeomorphism (a continuous bijection between compact Hausdorff spaces is a homeomorphism).

   2. Consider the equivalence relation on the unit square given by if or if . Prove that the quotient space is homeomorphic to .

      Let be the triangle with vertices and . The map given by is continuous and surjective. It satisfies iff therefore it descends to a continuous map (indeed a continuous bijection) . The is a homeomorphism (because a continuous bijection from a compact space to a Hausdorff space is a homeomorphism: is compact because it's the image of under a continuous map; is Hausdorff because it's a subspace of ). Finally, is homeomorphic to by (b.i).

3. Let be the closed unit disc, and let be the closed upper half plane. Consider the following four topological spaces: all of them equipped with the subspace topology from .
   1. Which two of the above four spaces are homeomorphic? Prove that they are indeed homeomorphic. [Hint: Use (b) (ii) to construct the desired homeomorphism.]

      Let be the homeomorphism constructed in part (b), and let be the point corresponding to the subset (note that is an equivalence class for the equivalence relation ∼). By (b.i), we have the following two homeomorphisms: and . The map induces a homeomorphism . The composite map induces a homeomorphism . Assembling all these maps, we get a homeomorphism

   2. Prove that no other two of the above four spaces are homeomorphic. [Hint: Which ones are locally compact?]

      We count the number of points that do not admit compact neighbourhoods: and admit no such points (they are locally compact). has two such points. has one such point.

1. Provide an example of an abstract simplicial complex whose topological realisation is homeomorphic to ℝ.

   Take as the vertex set, and as the set of simplices.

2. 1. Let be a topological space. Let be a set, and let be a surjective function.
      α) The map induces a bijection , for a certain equivalence relation ∼ on . Describe the equivalence relation ∼.

      The equivalence relation ∼ is given by declaring when .

      β) Describe the unique topology on which makes into a homeomorphism (where is equipped with the quotient topology).

      A subspace is open in this topology iff is open in .

   2. Define what it means for a continuous map between topological spaces to be a quotient map.

      is a quotient map if

   3. Let be the set of complex numbers of norm one, topologised as a subset of . Fix a real number, and consider the half-open interval . Is the map a quotient map? Justify your answer. [If the answer depends on the choice of , treat the different cases separately.]
      • This is not true when as, in that case, the map is a bijection. A bijective quotient map is a homeomorphism. But is compact whereas is not. So is not a quotient map.
      • This is true when . Given a subset whose preimage is open in , we must show that is open in . Pick a point . We will show that there exists an open arc containing , entirely contained inside . Let be the unique preimage of under such that . Since lies in the interior of and is open in , there exists such that . It follows that .

Declare if their images in are equal. The composite induces a continuous bijectionthat fits into a commutative triangle A subspace is open iff its preimage in is open (by definition of the quotient topology on ) iff its preimage in is open (by definition of the quotient topology on . Similarly, a subspace is open iff its preimage in is open. The bijective map induces a bijection between the open subsets of and the open subsets of . So is a homeomorphism.

The first space is the standard definition of . The second space can be viewed as an iterated quotientas in (c). The first quotient is homeomorphic to via the quotient map (a bijective continuous map between compact Hausdorff spaces is a homeomorphism). The equivalence relation on points of induces the equivalence relation on that identifies antipodal points of the boundary. The quotient of is another standard definition of . Explicitly, the embedding induces a homeomorphism (a bijective continuous map between compact Hausdorff spaces is a homeomorphism).
is isomorphic to the Klein bottle .