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Table of Contents

  1. 2.1.2
  2. 2.1.4
  3. 2.1.6
  4. 2.1.11
  5. 2.1.12
  6. 2.1.13
  7. 2.1.15
  8. 2.1.18
  9. 2.1.20
  10. 2.1.25
  11. 2.1.30
  12. 2.1.31

2.1.2 #

AT-2.1.2. Show that the Δ-complex obtained from Δ3 by performing the order-preserving edge identifications [𝑣0,𝑣1]∼[𝑣1,𝑣3] and [𝑣0,𝑣2]∼[𝑣2,𝑣3] deformation retracts onto a Klein bottle. Also, find other pairsof identifications of edges that produce Δ-complexes deformation retracting onto a torus, a 2-sphere, andℝP2.Solution by Jihyeon Kim (김지현) (simnalamburt)𝑣0𝑣1𝑣2𝑣3𝑣0𝑣1𝑣2𝑣3𝑣0𝑣1𝑣2𝑣3𝑣0𝑣1𝑣2𝑣3Figure 1: Deformation retract onto a Klein bottle

2.1.4 #

AT-2.1.4. Compute the simplicial homology groups of the triangular parachute obtained from Δ2 byidentifying its three vertices to a single point.Solution by finalchildThe complex has one 0-simplex.𝐻Δ0(𝑋)=ℤThe complex has three 1-simplex, which is all a cycle, where 𝑎+𝑏+𝑐 forms a boundary.𝐻Δ1(𝑋)=ℤ2The complex has one 2-simplex, which is not a cycle.𝐻Δ2(𝑋)=0The complex has no 𝑛-simplex where 𝑛≥3.𝐻Δ𝑛(𝑋)=0 (𝑛≥3)

2.1.6 #

AT-2.1.6. Compute the simplicial homology groups of the Δ-complex obtained from 𝑛+1 2-simplicesΔ20,⋯,Δ2𝑛 by identifying all three edges of Δ20 to a single edge, and for 𝑖>0 identifying the edges [𝑣0,𝑣1]and [𝑣1,𝑣2] of Δ2𝑖 to a single edge and the edge [𝑣0,𝑣2] to the edge [𝑣0,𝑣1] of Δ2𝑖−1.Solution by Jineon Baek (백진언) (xtalclr)Call the space 𝑋. Call the 2-simplex generators 𝑓0,…,𝑓𝑛. Call the edge [𝑣0,𝑣1] of Δ2𝑖 as 𝑒𝑖. Identifying allthe edges as told, we can see that 𝑒0,…,𝑒𝑛 are all the representators of the identifications done on the edges(that is, every edge is identified with exactly one of 𝑒0,…,𝑒𝑛). So they form the 1-simplex generators of 𝑋.The space 𝑋 has only one 0-simplex: all points of Δ20 are identified, then 𝑣0 of Δ2𝑖 is identified with 𝑣0 ofΔ2𝑖−1, then [𝑣0,𝑣1]=[𝑣1,𝑣2] on Δ2𝑖.With all this, we now have a complete understanding of the chain maps. Under 𝜕, each generator maps asfollows.•𝑓0 maps to 𝑒0−𝑒0+𝑒0=𝑒0•For 𝑖>0, 𝑓𝑖 maps to [𝑣0,𝑣1]−[𝑣0,𝑣2]+[𝑣1,𝑣2]=2𝑒𝑖−𝑒𝑖−1.•All edges map to zero as we have only one point.Now we compute the homology.•𝜕 on 𝐶2(𝑋) is injective and we have 𝐻2(𝑋)=0.•The homology 𝐻1(𝑋) is ⨁𝑖ℤ𝑒𝑖 factored out by the image of 𝜕:𝐶2(𝑋)→𝐶1(𝑋). This identifies 𝑒0=0and 2𝑒𝑖=𝑒𝑖−1. So we have 𝐻1(𝑋)≅ℤ/2𝑛ℤ. See the proof below.‣Define the map ℤ→𝐻1(𝑋) by setting 𝑧↦𝑧𝑒𝑛. The map is surjective: repeatedly use 𝑒𝑖−1=2𝑒𝑖 topush any element of 𝐻1(𝑋) to the form 𝑧𝑒𝑛. The kernel of the map contains 2𝑛 so now we have themap ℤ/2𝑛ℤ→𝐻1(𝑋). The inverse 𝐻1(𝑋)→ℤ/2𝑛ℤ is 𝑒𝑖↦2𝑛−𝑖 (it is routine to check that the mapsare well-defined inverse of each other; track the generators).•We have 𝐻0(𝑋)≅ℤ as 𝑋 is nonzero and connected. 𝐻𝑛(𝑋)=0 for all 𝑛≥3 by dimensionality.

2.1.11 #

AT-2.1.11. Show that if 𝐴 is a retract of 𝑋 then the map 𝐻𝑛(𝐴)→𝐻𝑛(𝑋) induced by the inclusion 𝐴 ⊂𝑋 is injective.Solution by finalchildThe retract is the left inverse of the inclusion. By functoriality, the map induced by the retract is the leftinverse of the map induced by the inclusion. Thus, the map induced by the inclusion is injective.∎

2.1.12 #

AT-2.1.12. Show that chain homotopy of chain maps is an equivalence relation.Solution by kiwiyouGiven two chain maps 𝑓♯,𝑔♯:𝐶𝑛(𝑋)→𝐶𝑛(𝑌), let’s say 𝑓♯∼𝑔♯ if there exists a chain homotopy 𝑃:𝐶𝑛(𝑋)→𝐶𝑛+1(𝑌).We want to show that ∼ is an equivalence relation.1. Reflexivity𝑓♯−𝑓♯=0=𝜕(0)+0=𝜕𝟎+𝟎𝜕∎2. Symmetry𝑓♯∼𝑔♯⟹∃𝑃:𝜕𝑃+𝑃𝜕=𝑔♯−𝑓♯⟹∃𝑃:𝜕(−𝑃)+(−𝑃)𝜕=𝑓♯−𝑔♯⟹𝑔♯∼𝑓♯∎3. TransitivityLet’s introduce another chain map ℎ♯:𝐶𝑛(𝑋)→𝐶𝑛+1(𝑌).𝑓♯∼𝑔♯∧𝑔♯∼ℎ♯⟹∃𝑃,𝑄:𝜕𝑃+𝑃𝜕=𝑔♯−𝑓♯∧𝜕𝑄+𝑄𝜕=ℎ♯−𝑔♯⟹∃𝑃,𝑄:𝜕(𝑃+𝑄)+(𝑃+𝑄)𝜕=ℎ♯−𝑓♯⟹𝑓♯∼ℎ♯∎

2.1.13 #

AT-2.1.13. Verify that 𝑓≅𝑔 implies 𝑓∗=𝑔∗ for induced homomorphisms of reduced homology groups.Solution by Jineon Baek (백진언) (xtalclr)We first parse the problem.•The maps 𝑓,𝑔:𝑋→𝑌 are continuous maps from topological space 𝑋 to 𝑌.•That 𝑓≅𝑔 means that there is a homotopy 𝐻:𝐼×𝑋→𝑌 from 𝑓 to 𝑔‣so that 𝐻 is continuous, and 𝐻(0,−)=𝑓, and 𝐻(1,−)=𝑔•The maps 𝑓∗,𝑔∗:̃𝐻𝑛(𝑋)→̃𝐻𝑛(𝑌) are induced homomorphisms of reduced homology groups (p113).‣We will call them ̃𝑓∗,̃𝑔∗ instead to be explicit that we are talking about reduced homology.‣Side note: the maps 𝑓#,𝑔# are induced homomorphisms on chains (p110).Recall that even for non-reduced homology groups, showing 𝑓∗=𝑔∗ was a nontrivial task (Theorem 2.10).The proof constructed the prism operator 𝑃:𝐶𝑛(𝑋)→𝐶𝑛+1(𝑌) so that it satisfies the relation𝜕𝑃=𝑔#−𝑓#−𝑃𝜕on every dimension 𝑛. Here we consider 𝑔#,𝑓# be on dimension 𝑛. Consequently, the 𝑃 on left-hand side is𝑃:𝐶𝑛(𝑋)→𝐶𝑛+1(𝑌) and the right-hand side is 𝑃:𝐶𝑛−1(𝑋)→𝐶𝑛(𝑌).Our job now is to modify the prism operator and the equation above for reduced cohomology. Call thecorresponding homomorphisms on reduced chains ̃𝑓#,̃𝑔# to avoid confusion.1.The only difference with 𝑓#,𝑔# and ̃𝑓#,̃𝑔# is that, at dimension −1, we have ̃𝑓#,̃𝑔#:ℤ→ℤ instead of𝑓#,𝑔#:0→0.2.So define ̃𝑃:̃𝐶𝑛(𝑋)→̃𝐶𝑛+1(𝑌) the same as 𝑃 for dimension 𝑛≠−1,−2 (you don’t have any differencein domain and range). For 𝑛=−2, you don’t have a choice but set ̃𝑃 to zero as ̃𝐶−2(𝑋)=0. We willpostpone deciding ̃𝑃 for dimension 𝑛=−1.3.By 2 only, the equation 𝜕̃𝑃=̃𝑔#−̃𝑓#−̃𝑃𝜕 is already satisfied for 𝑛≠0,−1 as it does not have anydifference with original 𝜕𝑃=𝑔#−𝑓#−𝑃𝜕.4.For 𝑛=0, we have 𝜕̃𝑃=𝜕𝑃=𝑔#−𝑓#−𝑃𝜕=̃𝑔#−̃𝑓#−𝑃𝜕 that we need to match with ̃𝑔#−̃𝑓#−̃𝑃𝜕. What we get is that 𝑃𝜕=̃𝑃𝜕 for dimension zero. As 𝑃:𝐶−1(𝑋)→𝐶0(𝑌) should be a zero map (as𝐶−1(𝑋)=0) a natural guess is to set ̃𝑃:𝐶−1(𝑋)→𝐶0(𝑌) to be the zero map too. Let us do so, and thisclears up proving the equation for 𝑛=0.5.It remains to show 𝜕̃𝑃=̃𝑔#−̃𝑓#−̃𝑃𝜕 for 𝑛=−1. The left-hand side is a zero map by definition (see 4above). ̃𝑓#,̃𝑔#:ℤ→ℤ are identity maps so they cancel out. It remains to verify ̃𝑃𝜕=0 but recall that,in step 2 above, we set ̃𝑃:̃𝐶−2(𝑋)→̃𝐶−1(𝑋) to be zero. So this equation is verified.With all this, we defined the prism operator ̃𝑃:̃𝐶𝑛(𝑋)→̃𝐶𝑛+1(𝑌) for reduced homology and verifiedthe equation 𝜕̃𝑃=̃𝑔#−̃𝑓#−̃𝑃𝜕. So there is a chain homotopy between ̃𝑓# and ̃𝑔#, and thus ̃𝑓∗=̃𝑔∗(Proposition 2.12, that chain-homotopic chain maps induce the same homomorphisms on homology).Note. Why bother with ̃𝑃 when you can do the casework using 𝐻𝑛(𝑋)=̃𝐻𝑛(𝑋) for 𝑛≠0 and 𝐻0(𝑋)≅̃𝐻0(𝑋)⊕ℤ for nonempty 𝑋? My line of thought was that (i) the proof above would be safe in terms ofnaturality (I have to admit I don’t know exactly what I am talking about) but I fear that such ‘exceptional’casework might lead to some faliure in preserving naturality, and that (ii) the casework would be quite messyconsidering cases of 𝑋 or 𝑌 being empty or not too. Feel free to prove me wrong and come up with a cleanerproof instead.

2.1.15 #

AT-2.1.15. For an exact sequence 𝐴→𝐵→𝐶→𝐷→𝐸 show that 𝐶=0 iff the map 𝐴→𝐵 is surjectiveand 𝐷→𝐸 is injective. Hence for a pair of spaces (𝑋,𝐴), the inclusion 𝐴↪𝑋 induces isomorphisms on allhomology groups iff 𝐻𝑛(𝑋,𝐴)=0 for all 𝑛.Solution by finalchild(a)For an exact sequence 𝐴→𝐵→𝐶→𝐷→𝐸 show that 𝐶=0 iff the map 𝐴→𝐵 is surjective and 𝐷→𝐸 is injective.⇒)img(𝐴→𝐵)=ker(𝐵→𝐶)=𝐵ker(𝐷→𝐸)=img(𝐶→𝐷)=0⇐)ker(𝐶→𝐷)=img(𝐵→𝐶)=𝐵/ker(𝐵→𝐶)=𝐵/img(𝐴→𝐵)=0img(𝐶→𝐷)=ker(𝐷→𝐸)=0Thus 𝐶=0(b)For a pair of spaces (𝑋,𝐴), the inclusion 𝐴↪𝑋 induces isomorphisms on all homology groups iff𝐻𝑛(𝑋,𝐴)=0 for all 𝑛.⇒)Consider the exact sequence …→𝐻𝑛+1(𝐴)→𝐻𝑛+1(𝑋)→𝐻𝑛+1(𝑋,𝐴)→𝐻𝑛(𝐴)→𝐻𝑛(𝑋)→…Since 𝐻𝑛(𝐴)→𝐻𝑛(𝑋) is surjective and injective for all 𝑛, it follows from (a) that 𝐻𝑛(𝑋,𝐴)=0 for all𝑛.⇐)Consider the exact sequence …→𝐻𝑛+1(𝐴)→𝐻𝑛+1(𝑋)→𝐻𝑛+1(𝑋,𝐴)→𝐻𝑛(𝐴)→𝐻𝑛(𝑋)→…Since for all 𝑛, 𝐻𝑛(𝑋,𝐴)=0, it follows from (a) that for all 𝑛, 𝐻𝑛(𝐴)→𝐻𝑛(𝑋) is surjective andinjective, thus an isomorphism.

2.1.18 #

AT-2.1.18. Show that for the subspace ℚ⊂ℝ, the relative homology group 𝐻1(ℝ,ℚ) is free abelian andfind a basis.Solution by Jihyeon Kim (김지현) (simnalamburt)𝐴⊂𝑋 를 만족하는 아무 두 위상공간 (𝐴,𝑋)에 대해, 아래 체인이 long exact sequence임을 Hatcher 115p 에서증명했다.⋯→𝐻𝑛(𝐴)→𝐻𝑛(𝑋)→𝐻𝑛(𝑋,𝐴)→𝐻𝑛−1(𝐴)→𝐻𝑛−1(𝑋)→⋯→𝐻0(𝑋,𝐴)→0이 체인에 𝑛≔1,𝐴≔ℚ,𝑋≔ℝ 를 대입하면 다음과 같다.⋯→𝐻1(ℚ)→𝐻1(ℝ)→𝐻1(ℝ,ℚ)→𝐻0(ℚ)→𝐻0(ℝ)→𝐻0(ℝ,ℚ)→0ℝ은 contractible 하므로 𝐻1(ℝ)=0,𝐻0(ℝ)=ℤ 이다. 따라서 위 체인은 다음과 같다.⋯→𝐻1(ℚ)→0→𝐻1(ℝ,ℚ)→𝐻0(ℚ)→ℤ→𝐻0(ℝ,ℚ)→0그리고 아래의 사실들에 의해,•ℚ는 0차원 동형사상으로 이루어진 무한개의 점으로 이루어져 있다.•path-connected인 standard n-simplex Δ𝑛를 totally-disconnected인 ℚ로 연속사상하는 것은 상수사상밖에없다.•singular n-simplex in ℚ의 정의는, standard n-simplex Δ𝑛를 ℚ로 보내는 연속사상이다.•𝐶𝑛(ℚ)는 singular n-simplices in ℚ의 집합을 기저로 갖의 free abelian group이다.아래가 성립한다:𝐶0(ℚ)=[𝑐]𝐶1(ℚ)=[𝑐,𝑐]𝐶2(ℚ)=[𝑐,𝑐,𝑐]𝜕1:𝐶1(ℚ)→𝐶0(ℚ)𝜕1([𝑐,𝑐])=[𝑐]−[𝑐]=0⋯ (zero map)Ker𝜕1=𝐶1(ℚ)Im𝜕1=0𝜕2:𝐶2(ℚ)→𝐶1(ℚ)𝜕2([𝑐,𝑐,𝑐])=[𝑐,𝑐]−[𝑐,𝑐]+[𝑐,𝑐]=[𝑐,𝑐]⋯ (constant map)Im𝜕2=𝐶1(ℚ)𝐻1(ℚ)=Ker𝜕1Im𝜕2=𝐶1ℚ𝐶1(ℚ)=0⋯ (trivial group)𝐻0(ℚ)≅⨁𝑞∈ℚℤ⋅[𝑞]⋯ (0th homology group counts path-connected components)이를 체인에 대입하면 다음과 같다.⋯→0→0→𝐻1(ℝ,ℚ)→⨁𝑞∈ℚℤ⋅[𝑞]→ℤ→𝐻0(ℝ,ℚ)→0여기서, 아래가 성립한다:let𝜕:𝐻1(ℝ,ℚ)→⨁𝑞∈ℚℤ⋅[𝑞]let𝑖∗:⨁𝑞∈ℚℤ⋅[𝑞]→ℤ𝑖∗=𝜆∑𝑞∈𝐹𝑛𝑞⋅[𝑞].∑𝑞∈𝐹𝑛𝑞⋯ (induced by the inclusion𝑖:ℚ↪ℝ)Ker𝜕=Im(0→𝐻1(ℝ,ℚ))=0⋯ (exactness, zero map)Im𝜕=𝐻1(ℝ,ℚ)Ker𝜕=𝐻1(ℝ,ℚ)⋯ (first isomorphism theorem)Im𝜕=Ker𝑖∗⋯ (exactness)따라서, 아래가 성립한다:𝐻1(ℝ,ℚ)≅Ker𝑖∗={∑𝑞∈𝐹𝑛𝑞⋅[𝑞]∈⨁𝑞∈ℚℤ⋅[𝑞]|∑𝑞∈𝐹𝑛𝑞=0}⋯ (definition of kernel)⊂{∑𝑞∈𝐹𝑛𝑞⋅[𝑞]∈⨁𝑞∈ℚℤ⋅[𝑞]}=⨁𝑞∈ℚℤ⋅[𝑞]⋯ (this is abelian group)𝐻1(ℝ,ℚ)는 abelian group의 subgroup이므로, 𝐻1(ℝ,ℚ)도 abelian이다.그리고, 아래 집합을 보면:𝐵≔{[𝑞]−[𝑞0]|𝑞∈ℚ\{𝑞0}}⋯ (for a fixed𝑞0∈ℚ)𝐵를 span하여 생성되는 부분군이 𝐻1(ℝ,ℚ) 전체가 됨을 알 수 있다:∑𝑞∈𝐹\𝑞0𝑛𝑞⋅([𝑞]−[𝑞0])=∑𝑞∈𝐹\𝑞0𝑛𝑞⋅[𝑞]−(((∑𝑞∈𝐹\𝑞0𝑛𝑞)))⋅[𝑞0]=∑𝑞∈𝐹\𝑞0𝑛𝑞⋅[𝑞]+𝑛𝑞0⋅[𝑞0]wherelet𝑛𝑞0≔−∑𝑞∈𝐹\𝑞0𝑛𝑞=∑𝑞∈𝐹𝑛𝑞⋅[𝑞]where∑𝑞∈𝐹𝑛𝑞=0𝐵는 선형독립이다. 𝐵에서 서로 다른 m개의 원소를 뽑아 선형결합하면:𝑘1([𝑞1]−[𝑞0])+𝑘2([𝑞2]−[𝑞0])+⋯+𝑘𝑚([𝑞𝑚]−[𝑞0])=(𝑘1[𝑞1]+𝑘2[𝑞2]+⋯+𝑘𝑚[𝑞𝑚])−(𝑘1+𝑘2+⋯+𝑘𝑚)[𝑞0]인데, 위 식이 0이 되는 유일한 방법은 𝑘1=𝑘2=⋯=𝑘𝑚=0이기 때문이다.따라서 𝐵는 𝐻1(ℝ,ℚ)의 기저다. Free abelian group의 정의는 기저를 갖는 abelian group이므로, 𝐻1(ℝ,ℚ)는free abelian group이다.

2.1.20 #

AT-2.1.20. Show that ̃𝐻𝑛(𝑋)≈̃𝐻𝑛+1(𝑆𝑋) for all 𝑛, where 𝑆𝑋 is the suspension of 𝑋. More generally,thinking of 𝑆𝑋 as the union of two cones 𝐶𝑋 with their bases identified, compute the reduced homologygroups of the union of any finite number of cones 𝐶𝑋 with their bases identified.Solution by kiwiyouTheorem. ∀𝑛:̃𝐻𝑛(𝑋)≈̃𝐻𝑛+1(𝑆𝑋)Proof. Let 𝑆𝑋 be the union of two cone 𝐶1𝑋 and 𝐶2𝑋 with their bases identified. Let their apexes be 𝑝1and 𝑝2 respectively.̃𝐻𝑛+1(𝑆𝑋)≃̃𝐻𝑛+1(𝑆𝑋,𝐶2𝑋) (∵long exact sequence)≃̃𝐻𝑛+1(𝑆𝑋∖{𝑝2},𝐶2𝑋∖{𝑝2}) (∵excision)≃̃𝐻𝑛(𝐶2𝑋∖{𝑝2}) (∵long exact sequence)≃̃𝐻𝑛(𝑋) (∵deformation retract) ∎.More generally, let 𝑈𝑘𝑋 be the union of 𝑘 cones with their bases identified.Theorem. ∀𝑛,𝑘:̃𝐻𝑛(𝑈𝑘𝑋)=(̃𝐻𝑛−1(𝑋))⊕(𝑘−1)Proof. Since 𝑈𝑘𝑋/𝐶𝑋≃(𝑆𝑋)∨(𝑘−1),̃𝐻𝑛(𝑈𝑘𝑋)≃̃𝐻𝑛(𝑈𝑘𝑋/𝐶𝑋) (∵long exact sequence)≃(̃𝐻𝑛(𝑆𝑋))⊕(𝑘−1)≃(̃𝐻𝑛−1(𝑋))⊕(𝑘−1) ∎.

2.1.25 #

AT-2.1.25. Find an explicit, noninductive formula for the barycentric subdivision operator 𝑆:𝐶𝑛(𝑋)→𝐶𝑛(𝑋).Solution by kiwiyouTheorem.𝑆Δ𝑛=∑𝜋∈Aut({0,1,…,𝑛})sgn(𝜋)[𝑏𝜋0,𝑏𝜋1,…,𝑏𝜋𝑛]where 𝑏𝜋𝑖 is the barycenter of [𝑣𝜋(𝑖),𝑣𝜋(𝑖+1),…,𝑣𝜋(𝑛)] and 𝑣𝑖 is the 𝑖-th vertex of Δ𝑛.Proof. Use induction. Let ̃𝜋∈Aut({0,1,…,𝑛+1}∖{𝑖}).𝑆Δ𝑛+1=∑𝑛+1𝑖=0(−1)𝑖[𝑏(Δ𝑛𝑖),𝑆Δ𝑛𝑖](inductive definition)=∑𝑛+1𝑖=0(−1)𝑖∑̃𝜋sgn(̃𝜋)[𝑏(Δ𝑛𝑖),𝑏̃𝜋0,𝑏̃𝜋1,…,𝑏̃𝜋𝑛](inductive hypothesis)=∑𝜋sgn(𝜋)[𝑏𝜋0,𝑏𝜋1,…,𝑏𝜋𝑛+1](sign of prepend)This explicit formula directly generalizes to general chains by the definition 𝑆𝜎=𝜎♯𝑆Δ𝑛.

2.1.30 #

AT-2.1.30. In each of the following commutative diagrams assume that all maps but one are isomorphisms.Show that the remaining map must be an isomorphism as well.Solution by finalchildTwo trivial properties of isomorphisms in a category: Inverse of an isomorphism is an isomorphism. Finitecomposition of isomorphisms is also an isomorphism.From the commutative diagrams, each map can be represented as a composition of other isomorphisms orinverses of other isomorphisms. Thus, the map is an isomorphism. ∎

2.1.31 #

AT-2.1.31. Using the notation of the five-lemma, give an example where the maps 𝛼, 𝛽, 𝛿, and 𝜀 are zerobut 𝛾 is nonzero. This can be done with short exact sequences in which all the groups are either ℤ or 0.Solution by finalchild↑0↑0↑id↑0↑0↑id↑0↑0↑0↑0↑id↑0↑000ℤℤ00ℤℤ00