§ Quotient topology

I watched this for shits and giggles. I don't know enough topology at all, but it's fun to watch arbitrary math videos.

§ Quotient topology: Defn, examples

  • Intended to formalise identifications.
Given space XX and equivalence relation \sim on XX, the quotient set X/X/\sim inherits a topology. Let q:XX/q : X \rightarrow X/\sim send point to equivalence class. Quotient topology is the most refined topology on X/X/\sim such that qq is continuous. That is, it has the most open sets for which this map is continuous.
  • More explicitly, a set UX/U \subset X/\sim (which is a collection of equivalence classes) is open iff q1(U)q^{-1}(U) is open in XX.
  • Even more explicitly, VX/V \subseteq X/\sim is open iff UV{xU:[x]V}U_V \equiv \{ x \in U : [x] \in V \}is open in XX.
  • Even more explicitly, we can write UvVvU \equiv \cup_{v \in V} v, because the elements of vv are equivalence classes.

§ Claim: quotient topology is a topology

  • The preimage of the empty set is the empty set, and thus is open.
  • The preimage of all equivalence classes is the full space, and thus open.
  • Preimage of union is union of preimages: iq1(Vi)\cup_i q^{-1}(V_i) extend hh to get a new homotopy HH: H0=idXH_0 = id_X and HtA=htH_t|A = h_t.

§ (X,A)(X, A) have HEP and AA is contractible, then XX/AX \simeq X/A

  • Pick q:XX/Aq: X \rightarrow X/A. We need another map such that their compositions are homotopic to the identities of XX and X/AX/A.
  • Define s:X/AXs: X/A \rightarrow X as a section of qq, given by s([a])a,s([x])xs([a]) \equiv a, s([x]) \equiv x. This is a section of qq since qs=idX/Aq \circ s = id_{X/A} (That is, ss maps entirely within the fibers of qq).
  • Consider st:Hts:X/AXs_t : H_t \circ s : X/A \rightarrow X. That is, lift from X/AX/A to XX using ss and then perform HtH_t on XX. We claim that The map (H1s)(H_1 \circ s) is the homotopy inverse of qq.
  • (1a) (H1s)q:XX(H_1 \circ s) \circ q : X \rightarrow X is equal to H1H_1, as H1(s(q(A)))=H1(s([a]))=H1(a)=a=H1(A)H_1(s(q(A))) = H_1(s([a])) = H_1(a) = a = H_1(A), and H1(s(q(x)))=H1(s([x]))=H1(x)H_1(s(q(x))) = H_1(s([x])) = H_1(x).
  • (1b) So we have (H1s)q=H1H0=idX(H_1 \circ s) \circ q = H_1 \simeq H_0 = id_X, as H0=idXH_0 = id_X is from defn, and H1H0H_1 \simeq H_0 is from homotopy. So we are done in this direction.
  • (2a) Consider q(H1s):XX/Aq \circ (H_1 \circ s) : X \rightarrow X/A. We wish to show that this is continuous. Let's show that it lifs to a continous map upstairs. So consider q(Hts)q:XX/Aq \circ (H_t \circ s) \circ q : X \rightarrow X/A. We claim that this is equal to qHtq \circ H_t, which is continuous as it is a composition of continuous maps.
  • This relationship is hopefully intuitive: q(Hts)qq \circ (H_t \circ s) \circ q asks us to treat all of AA as if it were aa before applying HtH_t. Since qq kills whatever HtH_t does after, and HtH_t guarantees to keep AA within AA, it's fine if we treat all of AA as just aa. qHtq \circ H_t asks us to treat AA as AA itself, and not aa. Since qq kills stuff anyway, we don't really care. The real crux of the argument is that qstabA=qstabAsqq \circ stab_A = q \circ stab_A \circ s \circ q where stabAstab_A is a map that stabilizes AA.
  • (2b) Consider (q(Hts)q)(A)=(qHts)([a])=(qHt)(a)(q \circ (H_t \circ s) \circ q)(A) = (q \circ H_t \circ s)([a]) = (q \circ H_t)(a) --- Since Ht(a)=ht(a)=aAH_t(a) = h_t(a) = a' \in A, we crush all data regardless of what happens. This is the same as the value (qHt)(A)=[a](q \circ H_t)(A) = [a] as Ht(A)AH_t(A) \subseteq A and q(A)=[a]q(A) = [a]. For the other set, we get (q(Hts)q)(x)=qHts([x])=qHt(x)(q \circ (H_t \circ s) \circ q)(x) = q \circ H_t \circ s([x]) = q \circ H_t(x) and hence we are done.
  • (2c) Now since q(Hts))q \circ (H_t \circ s)) is continuous, and that q(H0s):X/AX/A=idX/Aq \circ (H_0 \circ s) : X/A \rightarrow X/A = id_{X/A}, we are done since we can homotope from qH1sqH0s=idX/Aq \circ H_1 \circ s \simeq q \circ H_0 \circ s = id_{X/A}.
Slogan: Use HEP to find homotopy HH. Use H1sH_1 \circ s as inverse to quotient.