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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

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.