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 pairs of identifications of edges that produce Δ-complexes deformation retracting onto a torus, a 2-sphere, and ℝP2.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 by identifying 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-simplices Δ20,⋯,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 all the 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 of Δ2𝑖−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 as follows.•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=0 and 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 to push any element of 𝐻1(𝑋) to the form 𝑧𝑒𝑛. The kernel of the map contains 2𝑛 so now we have the map ℤ/2𝑛ℤ→𝐻1(𝑋). The inverse 𝐻1(𝑋)→ℤ/2𝑛ℤ is 𝑒𝑖↦2𝑛−𝑖 (it is routine to check that the maps are 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 left inverse 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∗:𝐻̃𝑛(𝑋)→𝐻̃𝑛(𝑌) are induced homomorphisms of reduced homology groups (p113).‣We will call them 𝑓̃∗,𝑔̃∗ 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 is 𝑃: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 the corresponding homomorphisms on reduced chains 𝑓̃#,𝑔̃# to avoid confusion.1.The only difference with 𝑓#,𝑔# and 𝑓̃#,𝑔̃# is that, at dimension −1, we have 𝑓̃#,𝑔̃#:ℤ→ℤ instead of 𝑓#,𝑔#:0→0.2.So define 𝑃̃:𝐶̃𝑛(𝑋)→𝐶̃𝑛+1(𝑌) the same as 𝑃 for dimension 𝑛≠−1,−2 (you don’t have any difference in domain and range). For 𝑛=−2, you don’t have a choice but set 𝑃̃ to zero as 𝐶̃−2(𝑋)=0. We will postpone deciding 𝑃̃ for dimension 𝑛=−1.3.By 2 only, the equation 𝜕𝑃̃=𝑔̃#−𝑓̃#−𝑃̃𝜕 is already satisfied for 𝑛≠0,−1 as it does not have any difference with original 𝜕𝑃=𝑔#−𝑓#−𝑃𝜕.4.For 𝑛=0, we have 𝜕𝑃̃=𝜕𝑃=𝑔#−𝑓#−𝑃𝜕=𝑔̃#−𝑓̃#−𝑃𝜕 that we need to match with 𝑔̃#−𝑓̃#−𝑃̃𝜕. What we get is that 𝑃𝜕=𝑃̃𝜕 for dimension zero. As 𝑃:𝐶−1(𝑋)→𝐶0(𝑌) should be a zero map (as 𝐶−1(𝑋)=0) a natural guess is to set 𝑃̃:𝐶−1(𝑋)→𝐶0(𝑌) to be the zero map too. Let us do so, and this clears up proving the equation for 𝑛=0.5.It remains to show 𝜕𝑃̃=𝑔̃#−𝑓̃#−𝑃̃𝜕 for 𝑛=−1. The left-hand side is a zero map by definition (see 4 above). 𝑓̃#,𝑔̃#:ℤ→ℤ are identity maps so they cancel out. It remains to verify 𝑃̃𝜕=0 but recall that, in step 2 above, we set 𝑃̃:𝐶̃−2(𝑋)→𝐶̃−1(𝑋) to be zero. So this equation is verified.With all this, we defined the prism operator 𝑃̃:𝐶̃𝑛(𝑋)→𝐶̃𝑛+1(𝑌) for reduced homology and verified the equation 𝜕𝑃̃=𝑔̃#−𝑓̃#−𝑃̃𝜕. So there is a chain homotopy between 𝑓̃# and 𝑔̃#, and thus 𝑓̃∗=𝑔̃∗ (Proposition 2.12, that chain-homotopic chain maps induce the same homomorphisms on homology).Note. Why bother with 𝑃̃ when you can do the casework using 𝐻𝑛(𝑋)=𝐻̃𝑛(𝑋) for 𝑛≠0 and 𝐻0(𝑋)≅𝐻̃0(𝑋)⊕ℤ for nonempty 𝑋? My line of thought was that (i) the proof above would be safe in terms of naturality (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 messy considering cases of 𝑋 or 𝑌 being empty or not too. Feel free to prove me wrong and come up with a cleaner proof 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 surjective and 𝐷→𝐸 is injective. Hence for a pair of spaces (𝑋,𝐴), the inclusion 𝐴↪𝑋 induces isomorphisms on all homology 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 iff 𝐻𝑛(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 all 𝑛.⇐)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 and injective, 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 and find 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 에서 증명했다.⋯→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 상수사상밖에 없다.•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)는 free abelian group이다.
2.1.20#AT-2.1.20. Show that 𝐻̃𝑛(𝑋)≈𝐻̃𝑛+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 homology groups of the union of any finite number of cones 𝐶𝑋 with their bases identified.Solution by kiwiyouTheorem. ∀𝑛:𝐻̃𝑛(𝑋)≈𝐻̃𝑛+1(𝑆𝑋)Proof. Let 𝑆𝑋 be the union of two cone 𝐶1𝑋 and 𝐶2𝑋 with their bases identified. Let their apexes be 𝑝1 and 𝑝2 respectively.𝐻̃𝑛+1(𝑆𝑋)≃𝐻̃𝑛+1(𝑆𝑋,𝐶2𝑋) (∵long exact sequence)≃𝐻̃𝑛+1(𝑆𝑋∖{𝑝2},𝐶2𝑋∖{𝑝2}) (∵excision)≃𝐻̃𝑛(𝐶2𝑋∖{𝑝2}) (∵long exact sequence)≃𝐻̃𝑛(𝑋) (∵deformation retract) ∎.More generally, let 𝑈𝑘𝑋 be the union of 𝑘 cones with their bases identified.Theorem. ∀𝑛,𝑘:𝐻̃𝑛(𝑈𝑘𝑋)=(𝐻̃𝑛−1(𝑋))⊕(𝑘−1)Proof. Since 𝑈𝑘𝑋/𝐶𝑋≃(𝑆𝑋)∨(𝑘−1),𝐻̃𝑛(𝑈𝑘𝑋)≃𝐻̃𝑛(𝑈𝑘𝑋/𝐶𝑋) (∵long exact sequence)≃(𝐻̃𝑛(𝑆𝑋))⊕(𝑘−1)≃(𝐻̃𝑛−1(𝑋))⊕(𝑘−1) ∎.
2.1.22#AT-2.1.22. Prove by induction on dimension the following facts about the homology of a finite-dimensional CW complex 𝑋, using the observation that 𝑋𝑛/𝑋𝑛−1 is a wedge sum of 𝑛-spheres:(a)If 𝑋 has dimension 𝑛 then 𝐻𝑖(𝑋)=0 for 𝑖>𝑛 and 𝐻𝑛(𝑋) is free.(b)𝐻𝑛(𝑋) is free with basis in bijective correspondence with the 𝑛-cells if there are no cells of dimension 𝑛−1 or 𝑛+1.(c)If 𝑋 has 𝑘 𝑛-cells, then 𝐻𝑛(𝑋) is generated by at most 𝑘 elements.Solution by kiwiyouLemma 1. For any integer 𝑛>0, 𝑋𝑛/𝑋𝑛−1=⋁𝛼𝑆𝑛𝛼.Proof. 𝑋𝑛 is the quotient space of 𝑋𝑛−1∐𝛼𝐷𝑛𝛼 under the identifications 𝑥∼𝜑𝛼(𝑥) for each 𝑥∈𝜕𝐷𝑛𝛼, where 𝜑𝛼:𝑆𝑛−1→𝑋𝑛−1. In 𝑋𝑛/𝑋𝑛−1, 𝑋𝑛−1 collapses into a single point, namely 𝑝. Then 𝜕𝐷𝑛𝛼 is all identified with 𝑝, which forms ⋁𝛼𝑆𝑛𝛼.Theorem (a). If 𝑋 has dimension 𝑛 then 𝐻𝑖(𝑋)=0 for 𝑖>𝑛 and 𝐻𝑛(𝑋) is free.Proof. It is trivial for 𝑛=0 (base case). Inductively suppose it holds for 𝑛<𝑘. Look at the long exact sequence for 𝑖>𝑘:⋯→𝐻𝑖(𝑋𝑘−1)→𝐻𝑖(𝑋𝑘)=𝐻𝑖(𝑋)→𝐻𝑖(𝑋𝑘,𝑋𝑘−1)→⋯Since (𝑋𝑘,𝑋𝑘−1) is a good pair, and by the inductive hypothesis, it becomes:⋯→0→𝐻𝑖(𝑋)→𝐻̃𝑖(𝑋𝑘/𝑋𝑘−1)→⋯Note that 𝐻̃𝑖(𝑋𝑘/𝑋𝑘−1)=𝐻̃𝑖(⋁𝛼𝑆𝑘𝛼)=⨁𝛼𝐻̃𝑖(𝑆𝑘)=0. Therefore, 𝐻𝑖(𝑋)=0 if 𝑖>𝑘. Now look at the long exact sequence for 𝑖=𝑘:⋯→0→𝐻𝑘(𝑋)→𝑞∗𝐻̃𝑘(𝑋𝑘/𝑋𝑘−1)→⋯Note that 𝐻̃𝑘(𝑋𝑘/𝑋𝑘−1)=𝐻̃𝑘(⋁𝛼𝑆𝑘𝛼)=ℤ𝛼. Exactness implies ker𝑞∗=0, which means 𝑞∗ is an inclusion. 𝐻𝑘(𝑋) is thus an abelian subgroup of free abelian group ℤ𝛼, and it must also be free. ∎Theorem (b). 𝐻𝑛(𝑋) is free with basis in bijective correspondence with the 𝑛-cells if there are no cells of dimension 𝑛−1 or 𝑛+1.Proof. Look at the long exact sequence when 𝑘≥𝑛+1:⋯→𝐻̃𝑛+1(𝑋𝑘+1/𝑋𝑘)→𝐻𝑛(𝑋𝑘)→𝐻𝑛(𝑋𝑘+1)→𝐻̃𝑛(𝑋𝑘+1/𝑋𝑘)→⋯Note that 𝐻̃𝑖(𝑋𝑘+1/𝑋𝑘)=⨁𝛼𝐻̃𝑖(𝑆𝑘+1𝛼)=0 when 𝑖≤𝑛+1<𝑘+1. Exactness implies 𝐻𝑛(𝑋𝑘)≅𝐻𝑛(𝑋𝑘+1). So if there are no cells of dimension 𝑛+1, we have infinite isomorphic chain starting with 𝐻𝑛(𝑋𝑛)=𝐻𝑛(𝑋𝑛+1):𝐻𝑛(𝑋𝑛)=𝐻𝑛(𝑋𝑛+1)≅𝐻𝑛(𝑋𝑛+2)≅⋯Since 𝑋 is finite-dimensional, this chain has a fixed point 𝐻𝑛(𝑋), and it is indeed isomorphic to 𝐻𝑛(𝑋𝑛). Now to find out basis, look at this long exact sequence:⋯→𝐻𝑛(𝑋𝑛−1)→𝐻𝑛(𝑋𝑛)→𝐻̃𝑛(𝑋𝑛/𝑋𝑛−1)→𝐻𝑛−1(𝑋𝑛−1)→⋯If there are no cells of dimension 𝑛−1 then 𝑋𝑛−1=𝑋𝑛−2. By Theorem (a), 𝐻𝑛(𝑋𝑛−1)=𝐻𝑛−1(𝑋𝑛−1)=0. We get final isomorphism:⋯→0→𝐻𝑛(𝑋𝑛)≅ℤ𝛼→0→⋯Here 𝛼 is from 𝑛-cells {𝑒𝑛𝛼}. Therefore, 𝐻𝑛(𝑋)≅𝐻𝑛(𝑋𝑛) is free with basis in bijective correspondence with the 𝑛-cells. ∎Theorem (c). If 𝑋 has 𝑘 𝑛-cells, then 𝐻𝑛(𝑋) is generated by at most 𝑘 elements.Recall the isomorphic chain from Theorem (b):𝐻𝑛(𝑋𝑛+1)≅𝐻𝑛(𝑋𝑛+2)≅⋯≅𝐻𝑛(𝑋)We don’t have 𝐻𝑛(𝑋𝑛)=𝐻𝑛(𝑋𝑛+1) this time. Instead look at another exact sequence:⋯→𝐻𝑛(𝑋𝑛)→𝑖∗𝐻𝑛(𝑋𝑛+1)→𝐻̃𝑛(𝑋𝑛+1/𝑋𝑛)→⋯Again from 𝐻̃𝑛(𝑋𝑛+1/𝑋𝑛)=0, im𝑖∗=𝐻𝑛(𝑋𝑛+1), thus 𝐻𝑛(𝑋) is abelian subgroup of 𝐻𝑛(𝑋𝑛). If 𝑋 has 𝑘 𝑛-cells, 𝐻̃𝑛(𝑋𝑛/𝑋𝑛−1)=ℤ𝑘. Same with the proof of Theorem (a), 𝐻𝑛(𝑋𝑛) is abelian subgroup of ℤ𝑘. Therefore 𝐻𝑛(𝑋) is generated by at most 𝑘 elements. ∎
2.1.24#AT-2.1.24. Show that each 𝑛-simplex in the barycentric subdivision Δ𝑛 is defined by 𝑛 inequalities 𝑡𝑖0≤𝑡𝑖1≤⋯≤𝑡𝑖𝑛 in its barycentric coordinates, where (𝑖0,𝑖1,⋯,𝑖𝑛) is a permutation of (0,1,⋯,𝑛).Solution by Jihyeon Kim (김지현) (simnalamburt)Δ𝑛 안의 한 점의 barycentric coordinates를 (𝑡0,𝑡1,…,𝑡𝑛)라 하자. 이들은 아래를 만족한다.𝑡𝑖≥0(𝑖=0,1,…,𝑛)∑𝑖𝑡𝑖=1(0,1,…,𝑛)의 순열중 임의로 하나를 골라 (𝑖0,𝑖1,…,𝑖𝑛)라 할 때, 𝑅을 아래와 같이 정의하겠다:𝑅≔{(𝑡0,𝑡1,…,𝑡𝑛)∈Δ𝑛|𝑡𝑖0≤𝑡𝑖1≤…≤𝑡𝑖𝑛}𝑅이 문제의 “𝑛 inequalities 𝑡𝑖0≤𝑡𝑖1≤⋯≤𝑡𝑖𝑛 in its barycentric coordinates, where (𝑖0,𝑖1,⋯,𝑖𝑛) is a permutation of (0,1,⋯,𝑛)”에 해당한다. 우리의 목표는 𝑅이 “each 𝑛-simplex in the barycentric subdivision Δ𝑛”임을 보이는 것이다.점 𝑝𝑚을 아래와 같이 정의하자:𝑝𝑚∈Δ𝑛(0≤𝑚≤𝑛)(𝑝𝑚)𝑖𝑗≔(0,0,…,0⏟𝑚,1𝑛−𝑚+1,1𝑛−𝑚+1,…,1𝑛−𝑚+1⏟𝑛−𝑚+1)𝑝𝑚은 면 [𝑖𝑚,…,𝑖𝑛]의 barycenter이다.이때 아래의 부등식 체인이 성립하는데,Δ𝑛=[𝑖0,𝑖1,…,𝑖𝑛]⊃[𝑖1,…,𝑖𝑛]⊃…⊃[𝑖𝑚,…,𝑖𝑛]⊃…⊃[𝑖𝑛]위 체인의 각 면의 barycenter는 각각 𝑝0,𝑝1,…,𝑝𝑚,…,𝑝𝑛이다. 따라서, Δ𝑛의 barycentric subdivision의 𝑛-simplex 중 현재 순열 (𝑖0,𝑖1,…,𝑖𝑛)에 대응하는 것은 [𝑝0,𝑝1,…,𝑝𝑛]이다. 이 simplex를 𝜎≔[𝑝0,𝑝1,…,𝑝𝑛]라 하자.이제 우리의 목표는 𝑅=𝜎임을 보이는 것이다.i) 𝜎⊂𝑅𝑝𝑚의 정의상, 𝑝𝑚은 𝑝𝑚∈Δ𝑛,(𝑝𝑚)𝑖0≤(𝑝𝑚)𝑖1≤…≤(𝑝𝑚)𝑖𝑛을 만족한다.0≤0≤…≤0≤1𝑛−𝑚+1≤1𝑛−𝑚+1≤…≤1𝑛−𝑚+1⇒(𝑝𝑚)𝑖0≤(𝑝𝑚)𝑖1≤…≤(𝑝𝑚)𝑖𝑛[𝑝0,𝑝1,…,𝑝𝑛]안의 임의의 점 𝑞는 𝑝0,𝑝1,…,𝑝𝑛의 convex combination이므로, 𝑞 역시 𝑞∈Δ𝑛,(𝑞)𝑖0≤…≤(𝑞)𝑖𝑛을 만족한다. 즉, 𝑞∈𝑅이다. 따라서, [𝑝0,𝑝1,…,𝑝𝑛]⊂𝑅이고, 𝜎⊂𝑅이다.ii) 𝜎⊃𝑅임의의 𝑡=(𝑡0,𝑡1,…,𝑡𝑛)∈𝑅에 대하여,𝑎0≔(𝑛+1)𝑡𝑖0𝑎𝑚≔(𝑛−𝑚+1)(𝑡𝑖𝑚−𝑡𝑖𝑚−1)(1≤𝑚≤𝑛)로 두자. 𝑡∈𝑅이므로 𝑡𝑖0≤𝑡𝑖1≤…≤𝑡𝑖𝑛이고, 따라서 각 𝑎𝑚≥0이다.또한,∑𝑛𝑚=0𝑎𝑚=(𝑛+1)𝑡𝑖0+∑𝑛𝑚=1(𝑛−𝑚+1)(𝑡𝑖𝑚−𝑡𝑖𝑚−1)=∑𝑛𝑗=0𝑡𝑖𝑗=∑𝑛𝑘=0𝑡𝑘=1이므로 (𝑎0,…,𝑎𝑛)은 barycentric coordinates이다.이제 𝑢≔∑𝑛𝑚=0𝑎𝑚𝑝𝑚라 두면, 𝑝𝑚의 정의에 의해 𝑚>𝑗이면 (𝑝𝑚)𝑖𝑗=0, 𝑚≤𝑗이면 (𝑝𝑚)𝑖𝑗=1𝑛−𝑚+1이다. 따라서 각 𝑗=0,1,…,𝑛에 대해𝑢𝑖𝑗=∑𝑗𝑚=0𝑎𝑚𝑛−𝑚+1=𝑡𝑖0+∑𝑗𝑚=1(𝑡𝑖𝑚−𝑡𝑖𝑚−1)=𝑡𝑖𝑗이다. 따라서 𝑢와 𝑡는 모든 barycentric coordinate가 같으므로 𝑢=𝑡이다. 즉, 𝑡=∑𝑛𝑚=0𝑎𝑚𝑝𝑚∈[𝑝0,𝑝1,…,𝑝𝑛]⊂𝜎.임의의 𝑡∈𝑅에 대해 𝑡∈𝜎이므로 𝑅⊂𝜎이다.결론적으로 𝜎⊂𝑅와 𝑅⊂𝜎가 모두 성립하므로 𝑅=𝜎.
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 𝜋̃∈Aut({0,1,…,𝑛+1}∖{𝑖}).𝑆Δ𝑛+1=∑𝑛+1𝑖=0(−1)𝑖[𝑏(Δ𝑛𝑖),𝑆Δ𝑛𝑖](inductive definition)=∑𝑛+1𝑖=0(−1)𝑖∑𝜋̃sgn(𝜋̃)[𝑏(Δ𝑛𝑖),𝑏𝜋̃0,𝑏𝜋̃1,…,𝑏𝜋̃𝑛](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. Finite composition of isomorphisms is also an isomorphism.From the commutative diagrams, each map can be represented as a composition of other isomorphisms or inverses 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 zero but 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