Split (1)

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space, i.e. the map from the set of covering spaces to the subgroups of π1 is surjective. Now we want to prove injectivity. To do so, we need a generalization of the homotopy lifting lemma. Suppose we have path-connected spaces (Y, y0), (X, x0) and ( ˜X, ˜x0), with f : (Y, y0) → (X, x0) a continuous map, p : ( ˜X, ˜x0...
follows: Given a y ∈ Y, there is some path αy : y0 y. Then f maps this to βy : x0 f (y) in X. By path lifting, this path lifts uniquely to ˜βy in ˜X. Then we set ˜f (y) = ˜βy(1). Note that if ˜f exists, then this must be what ˜f sends y to. What we need to show is that this is well-deﬁned. 35 3 Covering spaces II Alge...
� ˜V. whose end point is p\|−1 ˜V (f (z)) ∈ ˜V. So it follows that ˜f (z) = p\|−1 ˜V Now we prove that every subgroup of π1 comes from exactly one covering space. What this statement properly means is made precise in the following proposition: Proposition. Let (X, x0), ( ˜X1, ˜x1), ( ˜X2, ˜x2) be path-connected based spa...
x1) ˜p1 ( ˜X2, ˜x2) (X, x0) p2 p1 Now what we would like to do is to forget about the basepoints. What happens when we change base points? Recall that the eﬀect of changing base points is that we will conjugate our group. This doesn’t actually change the group itself, but if we are talking about subgroups, conjugation ...
−1b−1, or aa−1aaaaabbbb, etc. We are usually lazy and write aa−1aaaaabbbb as aa−1a5b4. When we see things like aa−1, we would want to cancel them. This is called elementary reduction. Deﬁnition (Elementary reduction). An elementary reduction takes a word usαs−1 α v and gives uv, or turns us−1 α sαv into uv. Since each ...
make it clear that the free group F (S) of any set S exists. We will state this deﬁnition as a lemma. Lemma. If G is a group and φ : S → G is a set map, then there exists a unique homomorphism f : F (S) → G such that the following diagram commutes: F (S) S f φ G where the arrow not labeled is the natural inclusion map...
(up to isomorphism), by taking G = F (S) and using the uniqueness properties. 39 4 Some group theory II Algebraic Topology Deﬁnition (Presentation of a group). Let S be a set, and let R ⊆ F (S) be any subset. We denote by R the normal closure of R, i.e. the smallest normal subgroup of F (S) containing R. This can be g...
. Example. a, b \| b ∼= a ∼= Z. Example. With a bit of work, we can show that a, b \| ab−3, ba−2 = Z/5. 4.2 Another view of free groups Recall that we have not yet properly deﬁned free groups, since we did not show that multiplication is well-deﬁned. We are now going to do this using topology. Again let S be a set. For t...
Notice that every word w ∈ S∗ denotes a unique “edge path” in ˜X starting at ˜x0, where an edge path is a sequence of oriented edges ˜e1, · · ·, ˜en such that the “origin” of ˜ei+1 is equal to the “terminus” of ˜ei. For example, the following path corresponds to w = abb−1b−1ba−1b−1. ˜x0 We can note a few things: (i) ˜...
�1(X, x0) are the same, since they are just concatenating words or paths. So this bijection identiﬁes the two group structures. So this induces an isomorphism F (S) ∼= π1(X, x0). 4.3 Free products with amalgamation We managed to compute the fundamental group of the circle. But we want to ﬁnd the fundamental group of mo...
seen to characterize G1 ∗ G2 up to isomorphism. Again, we have a deﬁnition in terms of a concrete construction of the group, without making it clear this is well-deﬁned; then we have a universal property 43 4 Some group theory II Algebraic Topology that makes it clear this is well-deﬁned, but not clear that the object...
44 5 Seifert-van Kampen theorem II Algebraic Topology 5 Seifert-van Kampen theorem 5.1 Seifert-van Kampen theorem The Seifert-van Kampen theorem is the theorem that tells us what happens when we glue spaces together. Here we let X = A ∪ B, where A, B, A ∩ B are path-connected. A x0 B We pick a basepoint x0 ∈ A ∩ B for...
. We want to ﬁnd π1(Sn). The idea is to write Sn as a union of two open sets. We let n = e1 ∈ Sn ⊆ Rn+1 be the North pole, and s = −e1 be the South pole. We let A = Sn \ {n}, 45 5 Seifert-van Kampen theorem II Algebraic Topology and B = Sn \ {s}. By stereographic projection, we know that A, B ∼= Rn. The hard part is to...
ive. Then the above proof strategy works. However, using Seifert-van Kampen is much neater. Example (RPn). Recall that RPn ∼= Sn/{± id}, and the quotient map Sn → RPn is a covering map. Now that we have proved that Sn is simply connected, we know that Sn is a universal cover of RPn. For any x0 ∈ RPn, we have a bijectio...
∼= Z ∗ 1 Z ∼= Z ∗ Z ∼= F2, where F2 is just F (S) for \|S\| = 2. We can see that Z ∗ Z ∼= F2 by showing that they satisfy the same universal property. Note that we had already ﬁgured this out when we studied the free group, where we realized F2 is the fundamental group of this thing. More generally, as long as x0, y0 in...
loop labelled a. a a a ˜x0 Note that ab−1 ∈ ker φ. So ab−1 gives a loop. So b goes in the same direction as a: b a ˜x0 b a a b 48 5 Seifert-van Kampen theorem II Algebraic Topology This is our covering space. This is a fun game to play at home: (i) Pick a group G (ﬁnite groups are recommended). (ii) Then pick α, β ∈ G...
, pick an arbitrary y1 ∈ A ∩ B. Since Dn is path connected, we have a path γ : y1 y0, and we can use this to recover the fundamental groups based at y0. Now Seifert-van Kampen theorem says π1(X ∪f Dn, y1) ∼= π1(A, y1) ∗ π1(A∩B,y1) π1(B, y1). Since B is just a disk, and A ∩ B is simply connected (n ≥ 3 implies Sn−1 is s...
π1(A ∩ B, y1) ∼= π1(X, x0)/f. In summary, we have π1(X ∪f Dn) = π1(X) n ≥ 3 π1(X)/f n = 2 This is a useful result, since this is how we build up cell complexes. If we want to compute the fundamental groups, we can just work up to the two-cells, and know that the higher-dimensional cells do not have any eﬀect. Moreover,...
) X such that π1(X) ∼= S \| R. There really isn’t anything that requires that ﬁniteness in this proof, but ﬁniteness makes us feel more comfortable. Proof. Let S = {a1, · · ·, am} and R = {r1, · · ·, rn}. We start with a single point, and get our X (1) by adding a loop about the point for each ai ∈ S. We then get our 2-...
Kampen theorem, but usually easier to apply, since we no longer have to “fatten up” our A and B to make them open. Proof. Pick open neighbourhoods A ∩ B ⊆ U ⊆ A and A ∩ B ⊆ V ⊆ B that strongly deformation retract to A ∩ B. Let U be such that U retracts to A ∩ B. Since U retracts to A, it follows that U is path connect...
a surface. Here we adopt the following deﬁnition: 52 5 Seifert-van Kampen theorem II Algebraic Topology Deﬁnition (Surface). A surface is a Hausdorﬀ topological space such that every point has a neighbourhood U that is homeomorphic to R2. Some people like C more that R, and sometimes they put C2 instead of R2 in the d...
dashed line. This would give two tori with a hole, where the boundary of the holes are just the dashed line. Then gluing back the dashed lines would give back our orientable surface with genus 2. In general, to produce Σg, we produce a polygon with 4g sides. Then we get π1Σg = a1, b1, · · ·, ag, bg \| a1b1a−1 1 b−1 1 ·...
for m = 1. This is just a fancy way of saying that R \ {0} is disconnected while Rm \ {0} is not for m = 1. We can just add 1 to n, and add 1 to our subscript. If Rm ∼= R2, then we have Rm \ {0} ∼= R2 \ {0} S1. We know that π1(S1) ∼= Z, while π1(Rm \ {0}) ∼= π1(Sm−1) ∼= 1 unless m = 2. The obvious thing to do is to cr...
is relatively more intuitive and can be computed directly. The drawback is that we will have to restrict to a particular kind of space, known as simplicial complexes. This is not a very serious restriction per se, since many spaces like spheres are indeed simplicial complexes. However, the deﬁnition of simplicial homo...
= 0, n i=0 ti = 0. t0 = − n i=1 ti. Then the ﬁrst equation reads 0 = − n i=1 ti a0 + t1a1 + · · · + tnan = n i=1 ti(ai − a0). So linear independence implies all ti = 0. 56 6 Simplicial complexes II Algebraic Topology The relevance is that these can be used to deﬁne simplices (which are simple, as opposed to complexes)...
e0, · · ·, en} in Rn+1. For example, when n = 2, we get the following: 57 6 Simplicial complexes II Algebraic Topology We will now glue simplices together to build complexes, or simplicial com- plexes. Deﬁnition. A (geometric) simplicial complex is a ﬁnite set K of simplices in Rn such that (i) If σ ∈ K and τ is a face...
. The boundary ∂σ is homeomorphic to Sn−1 (e.g. the boundary of a (solid) triangle is the boundary of the triangle, which is also a circle). This is called the simplicial (n − 1)-sphere. We can also triangulate our Sn in a diﬀerent way: Example. In Rn+1, consider the simplices ±e0, · · ·, ±en for each possible combinat...
a2)} also spans a 1-simplex because we are treating the collection of three vertices as a set, and not a simplex. f (a0), f (a1) f (a2) Finally, we can also do the following map: f (a0), f (a1) f (a2) The following lemma is obvious, but we will need it later on. Lemma. If K is a simplicial complex, then every point x ∈...
.e. StK(x) = ˚σ. x∈σ∈K The link of x, written LkK(x), is the union of all those simplices that do not contain x, but are faces of a simplex that does contain x. Deﬁnition (Simplicial approximation). Let f : \|K\| → \|L\| be a continuous map between the polyhedra. A function g : VK → VL is a simplicial approximation to f if...
need to check that im H ⊆ \|L\|. But we know that both \|g\|(x) and f (x) live in τ and τ is convex. It thus follows that H(x × I) ⊆ τ ⊆ \|L\|. To prove the last part, it suﬃces to show that every simplicial approximation to a simplicial map must be the map itself. Then the homotopy is rel M by the construction above. This ...
ˆσn : σi ∈ K and σ0 < σ1 < · · · < σn}. If you stare at this long enough, you will realize this is exactly what we have drawn above. The rth barycentric subdivision K (r) is deﬁned inductively as the barycentric subdivision of the r − 1th barycentric subdivision, i.e. K (r) = (K (r−1)). Proposition. \|K\| = \|K \| and K re...
large our mesh is: Lemma. Let dim K = n, then µ(K (r)) ≤ n r n + 1 µ(K). The key point is that as r → ∞, the mesh goes to zero. So indeed we can make our barycentric subdivisions ﬁner and ﬁner. The proof is purely technical and omitted. Now we can prove the simplicial approximation theorem. Proof of simplicial approxi...
now, we will forget about simplicial approximations and related ﬂuﬀ, but just note that it is ﬁne to assume everything can be considered to be simplicial. Instead, we are going to use this framework to move on and deﬁne some new invariants of simplicial complexes K, known as Hn(K). These are analogous to π0, π1, · · ·...
�, and then ¯σ denotes the same simplex with the opposite orientation. Example. As oriented 2-simplices, (v0, v1, v2) and (v1, v2, v0) are equal, but they are diﬀerent from (v2, v1, v0). We can imagine the two diﬀerent orientations of the simplices as follows: v1 v2 v0 v2 v0 v1 One and two dimensions are the dimensions...
meaningfully talk about Cn(K), but it’s just 0. Example. We can think of elements in the chain group C1(X) as “paths” in X. For example, we might have the following simplex: v1 σ1 σ2 v0 σ3 v2 σ6 Then the path v0 v1 v2 v0 v1 around the left triangle is represented by the member σ1 − σ2 + σ3 + σ1 = 2σ1 − σ2 + σ3. Of cou...
the homology groups as follows: Deﬁnition (Simplicial homology group Hn(K)). The nth simplicial homology group Hn(K) is deﬁned as Hn(K) = ker dn im dn+1. This is a nice, clean deﬁnition, but what does this mean geometrically? Somehow, Hk(K) describes all the “k-dimensional holes” in \|K\|. Since we are going to draw pic...
interpret its elements as k-dimensional “holes”. Example. Let K be the standard simplicial 1-sphere, i.e. we have the following in R3. e2 e1 e0 Our simplices are thus K = {e0, e1, e2, e0, e1, e1, e2, e2, e0}. Our chain groups are C0(K) = (e0), (e1), (e2) ∼= Z3 C1(K) = (e0, e1), (e1, e2), (e2, e0) ∼= Z3. All other chai...
1 im d2 ∼= ker d1. It is easy to see that in fact we have ker d1 = (e0, e1) + (e1, e2) + (e2, e0) ∼= Z. So we also have H1(K) ∼= Z. We see that this H1(K) is generated by precisely the single loop in the triangle. The fact that H1(K) is non-trivial means that we do indeed have a hole in the middle of the circle. Exampl...
. 7.2 Some homological algebra We will develop some formalism to help us compute homology groups Hk(K) in lots of examples. Firstly, we axiomatize the setup we had above. Deﬁnition (Chain complex and diﬀerentials). A chain complex C· is a sequence of abelian groups C0, C1, C2, · · · equipped with maps dn : Cn → Cn−1 su...
, but homotopies themselves are rather ﬂoppy. How can we deﬁne homotopies for chain complexes? It turns out we can have a completely algebraic deﬁnition for chain homotopies. Deﬁnition (Chain homotopy). A chain homotopy between chain maps f·, g· : C· → D· is a sequence of homomorphisms hn : Cn → Dn+1 such that gn − fn ...
otopy, i.e. H(∂σ × I) = hn−1 ◦ dn(σ). Now note that f (σ) and g(σ) have opposite orientations, so we get the result dn+1 ◦ hn(σ) = hn−1 ◦ dn(σ) + gn(σ) − fn(σ). Rearranging and dropping the σs, we get dn+1 ◦ hn − hn−1 ◦ dn = gn − fn. This looks remarkably like our deﬁnition for chain homotopies of maps, with the signs ...
� Bn(D). Hence gn(c) and fn(c) diﬀer by a boundary. So [gn(c)] − [fn(c)] = 0 in Hn(D), i.e. f∗(c) = g∗(c). The following statements are easy to check: Proposition. (i) Being chain-homotopic is an equivalence relation of chain maps. 72 7 Simplicial homology II Algebraic Topology (ii) If a· : A· → C· is a chain map and f...
a chain map f· : C·(K) → C·(L). Hence it also induces f∗ : Hn(K) → Hn(L). Proof. This is fairly obvious, except that simplicial maps are allowed to “squash” simplices, so f might send an n-simplex to an (n − 1)-simplex, which is not in Dn(L). We solve this problem by just killing these troublesome simplices. Let σ be ...
any simplex σ in LkK(v0) the simplex spanned by σ and v0. Details are left to the reader. Corollary. If ∆n is the standard n-simplex, and L consists of ∆n and all its faces, then Hk(L) = Proof. K is a cone (on any vertex). Z k = 0 k > 0 0 What we would really like is an example of non-trivial homology groups. Moreover...
invariant of Sn−1 for n > 2. We say this is just a “suggestion”, since the simplicial homology is deﬁned for simplicial complexes, and we are not guaranteed that if we put a diﬀerent simplicial complex on Sn−1, we will get the same homology groups. So the major technical obstacle we still need to overcome is to see th...
omorphisms fi−1 · · · Ai fi Ai+1 fi+1 Ai+2 fi+2 · · · is exact at Ai if We say it is exact if it is exact at every Ai. ker fi = im fi−1. Recall that we have seen something similar before. When we deﬁned the chain complexes, we had d2 = 0, i.e. im d ⊆ ker d. Here we are requiring exact equivalence, which is something ev...
· · where i∗ and j∗ are induced by i· and j·, and ∂∗ is a map we will deﬁne in the proof. Having an exact sequence is good, since if we know most of the terms in an exact sequence, we can ﬁgure out the remaining ones. To do this, we also need to understand the maps i∗, j∗ and ∂∗, but we don’t need to understand all, s...
∗−∗ H0(K) · · · 0 Here A ⊕ B is the direct sum of the two (abelian) groups, which may also be known as the Cartesian product. Note that unlike the Seifert-van Kampen theorem, this does not require the intersection L = M ∩ N to be (path) connected. This restriction was needed for Seifert-van Kampen since the fundamental...
∗ j∗ Hn(C) Hn−1(C) · · · where i∗ and j∗ are induced by i· and j·, and ∂∗ is a map we will deﬁne in the proof. 78 7 Simplicial homology II Algebraic Topology The method of proving this is sometimes known as “diagram chasing”, where we just “chase” around commutative diagrams to ﬁnd the elements we need. The idea of the...
is in the kernel of jn−1. Since the diagram is commutative, we know jn−1 ◦ dn(y) = dn ◦ jn(y) = dn(x) = 0, using the fact that x is a cycle. So dn(y) ∈ ker jn−1 = im in−1. Moreover, by exactness again, in−1 is injective. So there is a unique z ∈ An−1 such that in−1(z) = dn(y). We have now produced our z. We are not do...
z) = 0. (iii) (a) First, in the proof, suppose we picked a diﬀerent y such that jn(y) = jn(y) = x. Then jn(y − y) = 0. So y − y ∈ ker jn = im in. Let a ∈ An be such that in(a) = y − y. Then dn(y) = dn(y − y) + dn(y) = dn ◦ in(a) + dn(y) = in−1 ◦ dn(a) + dn(y). Hence when we pull back dn(y) and dn(y) to An−1, the result...
are chased along. It is even more beneﬁcial to attempt to prove this yourself. (a) im i∗ ⊆ ker j∗: This follows from the assumption that in ◦ jn = 0. (b) ker j∗ ⊆ im i∗: Let [b] ∈ Hn(B). Suppose j∗([b]) = 0. Then there is some c ∈ Cn+1 such that jn(b) = dn+1(c). By surjectivity of jn+1, there is some b ∈ Bn+1 such tha...
, and a ∈ An−1 such that in−1(a) = dn(b). By assumption, ∂∗([c]) = [a] = 0. So we know a is a boundary, say a = dn(a) for some a ∈ An. Then by commutativity we know dn(b) = dn ◦ in(a). In other words, dn(b − in(a)) = 0. So [b − in(a)] ∈ Hn(B). Moreover, j∗([b − in(a)]) = [jn(b) − jn ◦ in(a)] = [c]. So [c] ∈ im j∗. (e) ...
Moreover, we will show that they are homotopy invariants of the space, and that homotopic maps f g : \|K\| → \|L\| induce equal maps H∗(K) → H∗(L). Note that this is a lot to prove. At this point, we don’t even know arbitrary continuous maps induce any map on the homology. We only know simplicial maps do. We start with a ...
x if there are no repeats, 0 otherwise. We can now check by direct computation that this is indeed a chain homotopy. 82 7 Simplicial homology II Algebraic Topology Now we know that if f, g : K → L are both simplicial approximations to the same F, then they induce equal maps on homology. However, we do not yet know that...
Mayer-Vietoris sequence. Given a complicated simplicial complex K, let σ be a maximal dimensional simplex of K. We let L = K \ {σ} (note that L includes the boundary of σ). We let S = {σ and all its faces} ⊆ K and T = L ∩ S. We can do similarly for K, and let L, S, T be the corresponding barycentric subdivisions. We h...
−1 K,r, where s : K (r) → L is a simplicial approximation to f, and νK,r : Hn(K (r)) → Hn(K) is the isomorphism given by composing maps Hn(K (i)) → Hn(K (i−1)) induced by simplical approximations to the identity. Furthermore: (i) f∗ does not depend on the choice of r or s. (ii) If g : \|M \| → \|K\| is another continuous m...
that we get a simplicial approximation to both f and g. Proof. By the Lebesgue number lemma, there is an ε > 0 such that each ball of radius 2ε in \|L\| lies in some star StL(w). Now apply the Lebesgue number lemma again to {f −1(Bε(y))}y∈\|L\|, an open cover of \|K\|, and get δ > 0 such that for all x ∈ \|K\|, we have f (Bδ(...
∗ = g∗. This is good, since we know we can not only deal with spaces that are homeomorphic to complexes, but spaces that are homotopic to complexes. This is important, since all complexes are compact, but we would like to talk about non-compact spaces such as open balls as well. Deﬁnition (h-triangulation and homology ...
suitable machinery, we ﬁnd π1(S3 \ T ) ∼= a, b \| a3b−2. Staring at it hard enough, we can construct a surjection π1(S3 \ T ) → S3. This tells us π1(S3 \ T ) is non-abelian, and is certainly not Z. So we know U and T are genuinely diﬀerent knots. 7.6 Homology of spheres and applications Lemma. The sphere Sn−1 is triang...
0. So such a continuous retraction cannot exist. Note it is important that we can work with continuous maps directly, and not just their simplicial approximations. Otherwise, here we can only show that every simplicial approximation of f has a ﬁxed point, but this does not automatically entail that f itself has a ﬁxed...
prove our original proposition. Proposition. If n is even, then the antipodal map a id. Proof. We can directly compute that a∗x = (−1)n+1x. If n is even, then a∗ = −1, but id∗ = 1. So a id. 7.7 Homology of surfaces We want to study compact surfaces and their homology groups. To work with the simplicial homology, we ne...
�  Z n = 0 Z2g n = 1  n > 1 0 We now compute the homology groups of Σg. The Mayer-Vietoris sequence gives the following 0 H2(S1) H2(Fg−1) ⊕ H2(F1) H2(Σg) H1(S1) H1(Fg−1) ⊕ H1(F1) H1(Σg) H0(S1) H0(Fg−1) ⊕ H0(F1) H0(Σg) We can put in the terms we already know, and get 0 H2(Σg) Z Z2g H1(Σg) Z Z2 Z 0 0 By exactness, we...
cients instead. In the past, Cn(K) was an abelian group, or a Z-module. If we use rational coeﬃcients, since Q is a ﬁeld, this becomes a vector space, and we can use a lot of nice theorems about vector spaces, such as the rank-nullity theorem. Moreover, we can reasonably talk about things like the dimensions of these h...
Hn(Σg, Q) ∼=   Q Q2g  0 k = 0, 2 k = 1 otherwise. In this case, we have not lost any information because there was no torsion part of the homology groups. However, for the non-orientable surfaces, one can show that so Hk(En) =    Z k = 0 Zn−1 × Z/2 k = 1 0 otherwise, Hk(En; Q) =   Q Qn−1  0 k = 0 k = 1 ...
i≥0 (−1)i tr(f∗ : Hi(X; Q) → Hi(X; Q)). Why is this a generalization of the Euler characteristic? Just note that the trace of the identity map is the number of dimensions. So we have χ(X) = L(id). 92 7 Simplicial homology II Algebraic Topology Example. We have χ(Sn) = 2 n even 0 n odd We also have χ(Σg) = 2 − 2g, χ(En...
A = B A 0 C. What this allows us to do is to not look at the induced maps on homology, but just the maps on chain complexes. This makes our life much easier when it comes to computation. Corollary. Let f· : C·(K; Q) → C·(K; Q) be a chain map. Then (−1)i tr(fi : Ci(K) → Ci(K)) = i≥0 i≥0 (−1)i tr(f∗ : Hi(K) → Hi(K)), wi...
ned is the usual Euler characteristic, i.e. χ(X) = (−1)i number of i-simplices. Finally, we get to the important theorem of the section. i≥0 Theorem (Lefschetz ﬁxed point theorem). Let f : X → X be a continuous map from a triangulable space to itself. If L(f ) = 0, then f has a ﬁxed point. Proof. We prove the contrapos...
(gi ◦ si : Ci(K; Q) → Ci(K; Q)) Now note that si takes simplices of σ to sums of subsimplices of σ. So gi ◦ si takes every simplex oﬀ itself. So each diagonal terms of the matrix of gi ◦ si is 0! Hence the trace is L(\|g\|) = 0. Example. If X is any contractible polyhedron (e.g. a disk), then L(f ) = 1 for any f : X → X,...
can think of A as a subgroup of B, and C as the quotient B/A, by the ﬁrst isomorphism theorem. So any element of C can be represented by an element of B. We apply the boundary map to this representative, and then exactness shows that this must come from some element of A. We then check carefully that these is well-deﬁ...
z) = dn(y). We have now produced our z. We are not done. We have ∂∗[x] = [z] as our candidate deﬁnition, but we need to check many things: (i) We need to make sure ∂∗ is indeed a homomorphism. (ii) We need dn−1(z) = 0 so that [z] ∈ Hn−1(A); (iii) We need to check [z] is well-deﬁned, i.e. it does not depend on our choic...
−1 ◦ dn(a) + dn(y). Hence when we pull back dn(y) and dn(y) to An−1, the results diﬀer by the boundary dn(a), and hence produce the same homology class. 26 3 Four major tools of (co)homology III Algebraic Topology (b) Suppose [x] = [x]. We want to show that ∂∗[x] = ∂∗[x]. This time, we add a layer above. 0 0 0 An+1 in+...
qn+1, there is some b ∈ Bn+1 such that qn+1(b) = c. By commutativity, we know qn(b) = qn ◦ dn+1(b), i.e. qn(b − dn+1(b)) = 0. By exactness of the sequence, we know there is some a ∈ An such that in(a) = b − dn+1(b). Moreover, in−1 ◦ dn(a) = dn ◦ in(a) = dn(b − dn+1(b)) = 0, using the fact that b is a cycle. Since in−1...
So [b − in(a)] ∈ Hn(B). Moreover, q∗([b − in(a)]) = [qn(b) − qn ◦ in(a)] = [c]. So [c] ∈ im q∗. (e) im ∂∗ ⊆ ker i∗: Let [c] ∈ Hn(C). Let b ∈ Bn be such that qn(b) = c, and a ∈ An−1 be such that in(a) = dn(b). Then ∂∗([c]) = [a]. Then i∗([a]) = [in(a)] = [dn(b)] = 0. So i∗ ◦ ∂∗ = 0. (f) ker i∗ ⊆ im ∂∗: Let [a] ∈ Hn(A) ...
d) = 0. Since the sequence is exact, there is some c ∈ C such that h(c) = d. We are not yet done. We do not know that n(c) = c. All we know is that d(n(c)) = d(c). So d(c − n(c)) = 0. By exactness at C, we can ﬁnd some b 28 3 Four major tools of (co)homology III Algebraic Topology such that s(b) = n(c) − c. Since m was...
n + 1 simplex whose boundary is gn − fn, plus some terms arising from the boundary of c itself: gn(σ) gn(σ) Fn(σ) : dFn(σ) = + Fn−1(dσ) fn(σ) fn(σ) We will not attempt to justify the signs appearing in the deﬁnition; they are what are needed for it to work. The relevance of this deﬁnition is the following result: Lemm...
for n ≥ 0 such that d(Pn) = i1 − i0 − (−1)j([0, 1] × δj)#(Pn−1), n where j=0 i0 : δn = {0} × ∆n → [0, 1] × ∆n i1 : δn = {0} × ∆n → [0, 1] × ∆n and δj : ∆n−1 → ∆n is the inclusion of the jth face. These are “prisms” connecting the top and bottom face. Intuitively, the prism P2 looks like this: and the formula tells us ...
exists. We already had a picture of what it looks like, so we just need to ﬁnd a formula that represents it. We view [0, 1] × ∆n ∈ R × Rn+1. Write {v0, v1, · · ·, vn} for the vertices of {0} × ∆n ⊆ [1, 0] × ∆n, and {w0, · · ·, wn} for the corresponding vertices of {1} × ∆n. Now if {x0, x1, · · ·, xn+1} ⊆ {v0, · · ·, v...
3 Four major tools of (co)homology III Algebraic Topology It would be annoying if each choice of open cover gives a diﬀerent homology theory, because this would be too many homology theories to think about. The small simplices theorem says that the natural map H U ∗ (X) → H∗(X) is an isomorphism. Theorem (Small simpli...
with Hn(X). So we Now what does the boundary map ∂ : Hn(X) → Hn−1(A ∩ B) do? Suppose we have c ∈ Hn(X) represented by a cycle a + b ∈ C U n (X), with a supported in A and b supported in B. By the small simplices theorem, such a representative always exists. Then the proof of the snake lemma says that ∂([a + b]) is giv...
up, and one method to do so is barycentric subdivision. Given a 0-simplex {v0}, its barycentric subdivision is itself. If x = {x0, · · ·, xn} ⊆ Rn spans an n-simplex σ, we let bx = 1 n + 1 n i=0 xi be its barycenter. If we have a 1-simplex Then the barycentric subdivision is obtained as We can degenerately describe th...
ε(σ) · bn, 34 3 Four major tools of (co)homology III Algebraic Topology In total, we have i + Cone∆n where ci = 0 for i > 0, and c0(σ) = ε(σ)bn is a map C·(∆n) → C·(∆n). i−1d = id − c·, dCone∆n We now use this cone map to construct a barycentric subdivision map ρX n : Cn(X) → Cn(X), and insist that it is natural that ...
∈ C U n (X), then pX (c) ∈ C U Moreover, if c ∈ Cn(X), then there is some k such that (ρX n (X). n )k(c) ∈ C U n (X). Proof. The ﬁrst part is clear. For the second part, note that every chain is a ﬁnite sum of simplices. So we only have to check it for single simplices. We let σ be a simplex, and let V = {σ−1˚Uα} be a...
. We move on to (slightly) more interesting stuﬀ. The next few sections will all be slightly short, as we touch on various diﬀerent ideas. 36 4 Reduced homology III Algebraic Topology 4 Reduced homology Deﬁnition (Reduced homology). Let X be a space, and x0 ∈ X a basepoint. We deﬁne the reduced homology to be ˜H∗(X) = ...
�} = A/A → U/A H∗(X/A, A/A) H∗(X/A, U/A) is also an isomorphism. Now we have Hn(X, A) Hn(X/A, A/A) ∼ ∼ Hn(X, U ) excise A Hn(X \ A, U \ A) Hn(X/A, U/A) excise A/A Hn We now notice that X \ A = X vertical map is actually an isomorphism. So the result follows. A and. So the right-hand 37 5 Cell complexes III Algebraic To...
�ne X = X n n≥0 with the weak topology, namely that A ⊆ X is open if A ∩ X n is open in X n for all n. We write Φα : Dn α → X n for the obvious inclusion map. This is called the characteristic map for the cell eα. Deﬁnition (Finite-dimensional cell complex). If X = X n for some n, we say X is ﬁnite-dimensional. Deﬁniti...
Z = {xα : α ∈ In}, where each xα is the south pole of the sphere. To compute the homology of the wedge X n/X n−1, we then note that (Y, Z) is good, and so we have a long exact sequence α∈In Sn Hi(Z) Hi(Y ) ˜Hi(Y /Z) Hi−1(Z) Hi−1(Y ). Since Hi(Z) vanishes for i ≥ 1, the result then follows from the homology of the sphe...
1(X n−1, X n−2). We consider 0 0 Hn(X n+1) Hn(X n) ∂ qn Hn+1(X n+1, X n) dcell n+1 Hn(X n, X n−1) dcell n Hn−1(X n−1, X n−2) ∂ qn−1 Hn−1(X n−1) 0 Hn−1(X n) Referring to the above diagram, we see that n ◦ dcell dcell n+1 = qn−1 ◦ ∂ ◦ qn ◦ ∂ = 0, (X), dcell· ) is a since the middle ∂ ◦ qn is part of an exact sequence. So...
all free. We can similarly deﬁne cellular cohomology. Deﬁnition (Cellular cohomology). We deﬁne cellular cohomology by C n cell(X) = H n(X n, X n−1) and let dn cell be the composition H n(X n, X n−1) q∗ H n(X n) ∂ H n+1(X n+1, X n). This deﬁnes a cochain complex C· cell(X) with cohomology H ∗ cell(X) ∼= H ∗(X). H ∗ ce...
n−1 Sn−1 γ dcell n q Hn−1(X n−1, X n−2) excision ∼ ˜Hn−1(X n−1/X n−2) By the long exact sequence, the top left horizontal map is an isomorphism. Now let’s try to trace through the diagram. We can ﬁnd isomorphism 1 fαβ dαβ 1 eα dαγeγ dαγeγ So the degree of fαβ is indeed dαβ. Example. Let K be the Klein bottle. b π a v W...
Z Z So we have H0(K) = Z H1(K) = Z H2(K) = Z/2Z. Note that the second cohomology is not the dual of the second homology! However, if we forget where each factor is, and just add all the homology groups together, we get Z⊕Z⊕Z/2Z. Also, if we forget all the torsion components Z/2Z, then they are the same! 43 5 Cell comp...
even. Then we have · · · 2 Ze3 0 Ze2 2 Ze1 0 Ze0. What happens on the left end depends on whether n is even or odd. So we have Hi(RPn) =    Z i = 0 Z/2Z i odd, i < n 0 Z i even, 0 < i < n i = n is odd otherwise 0. We can immediately work out the cohomology too. We will just write out the answer: H i(RPn...
al) diﬀerential. We again set H n(X; A) = H n(C·(X; A)). We similarly deﬁne cellular cohomology. If A is in fact a commutative ring, then these are in fact R-modules. We call A the “coeﬃcients”, since a general member of C·(X; A) looks like nσs, where nσ ∈ A, σ : ∆n → X. We will usually take A = Z, Z/nZ or Q. Everythin...
n. Let n (X), which Zn = ker(dn : Cn → Cn−1) Bn = im(dn+1 : Cn+1 → Cn). We are now going to write down two short exact sequences. By deﬁnition of homology, we have 0 Bn Zn Hn(X; Z) 0. Also, the deﬁnition of Zn and Bn give us 0 Zn Cn Bn−1 0. We will now use the ﬁrst isomorphism theorem to know that the rank of the middl...
� Rk++1, and the restriction is given by σ\|[x0,...,xi](t0,..., ti) = σ. tjxj This is a bilinear map. Notation. We write H ∗(X; R) = H n(X; R). n≥0 This is the deﬁnition. We can try to establish some of its basic properties. We want to know how this interacts with the diﬀerential d with the cochains. The obvious answer ...
�er by −1. Then the remaining terms overlap in exactly 1 vertex, so we have ((dφ) ψ)(σ) + (−1)kφ (dψ)(σ) = (φ ψ)(dσ) = (d(φ ψ))(σ) as required. 47 8 Cup product III Algebraic Topology This is the most interesting thing about these things, because it tells us this gives a well-deﬁned map on cohomology. Corollary. The cu...
general for the cochains. So we would expect that this is rather annoying to prove. The proof relies on the following observation: Proposition. The cup product is natural, i.e. if f : X → Y is a map, and α, β ∈ H ∗(Y ; R), then f ∗(α β) = f ∗(α) f ∗(β). So f ∗ is a homomorphism of unital rings. Proof of previous propo...
φ ψ] = [ρ∗φ ρ∗ψ] = (−1)k[ρ∗(ψ φ)] = (−1)k[ψ φ] = (−1)klβ α. Now it remains to prove the claim. We have dρ(σ) = εn n (−1)jσ\|[vn,...,ˆvn−i,...,v0] i=0 n ρ(dσ) = ρ (−1)iσ\|[v0,...,ˆvi,....,vn] i=0 n = εn−1 j=0 (−1)jσ\|[vn,...,ˆvj,v0]. We now notice that εn−1(−1)n−i = εn(−1)i. So this is a chain map! We now deﬁne a chain hom...
v0, · · ·, vi, wn, · · ·, ˆwj, · · ·, vi]  .   The terms with j = i give (σ ◦ π)# i εn−i[v0, · · ·, vi−1, wn, · · ·, wi] (−1)n+1(−1)iεn−i[v0, · · ·, vi, wn, · · ·, wi+1] + i = (σ ◦ π)#(εn[wn, · · ·, w0] − [v0, · · ·, vn]) = ρ(σ) − σ The terms with j = i are precisely −F X n−1(dσ) as required. It is easy to see that...
R that vanishes on Ck(A). Then if σ ∈ Ck+(A) and ψ ∈ C (X; R), then (φ ψ)(σ) = φ(σ\|[v0,...,vk]) · ψ(σ\|[vk,...,vk+]). 50 8 Cup product III Algebraic Topology We now notice that [v0, · · ·, vk] ∈ Ck(A). So φ kills it, and this vanishes. So this is a term in H k+(X, A; R). You might ﬁnd it weird that the two factors of t...

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