Lecture Notes for Topology MMA100

Lecture notes for Topology MMA100

J A S, S-11

1 Simplicial Complexes

1.1 Aﬃne independence

N A collection of points v0, v1, . . . , vn in some Euclidean space R are affinely independent if the (affine hyper-) plane of least dimension containing them has dimension n. Synonymously, the points are in general or generic position. v3 This is equivalent to v1 − v0, v2 − v0, . . . , vn − v0 being linearly independent 3 v vectors (or similarly with any vi replacing v0). Thus e.g. four points in R 2 can be in generic position (but not five). ∑ v0 Another way∑ to express genericity is to say that tivi = 0, where ti are v1 scalars with ti = 0, only if all ti = 0. Yet another criterion is to use the embedding RN ,→ RN+1 given by adding R3 the scalar 1 as a first coordinate: x 7→ (1, x). Then v , v , . . . , v are in generic Figure 1: Four points in 0 1 n in general position position iff (1, v0), (1, v1),..., (1, vn) are linearly independent vectors.

Obviously, if v0, v1, . . . , vn are linearly independent as vectors they are also affinely independent as points (considering vectors as position vectors for points). ∑ ∑ An point p is an affine combination of the points v0, v1, . . . , vn if p = tivi, where ti = 1. If the points are in generic position, the affine coordinates ti are uniquely determined by p.

The (n-dimensional) plane spanned by points v0, v1, . . . , vn consists of all aﬃne combinations of them. 3 E.g. the (two-dimensional) plane spanned by three points v0, v1, v2 in R consist of all points x = t0v0 + t1v1 + t2v2, with t0 + t1 + t2 = 1, or diﬀerently put: x = v0 + t1(v1 − v0) + t2(v2 − v0), with no condition on the scalars. This is a standard parametric equation for the plane.

1.2 Convex closure

As you know a subset of RN is convex if whenever it contains two points p and q it contains the whole line segment between them. This is parametrized by (1 − s)p + sq, where s ∈ [0, 1]. It follows directly that the intersection of any number of convex sets is again convex. Consequently, for any subset A ⊂ RN there is a smallest convex set containing it: the intersection of all convex sets containing it. This is the convex closure, co(A) of A. ∑ RN ∈ An aﬃne combination∑ tipi of points p0, p1, . . . , pn in , is a convex combination if all ti [0, 1] (in addition to ti = 1).

An easy induction on n shows that if a0, a1, . . . , an are points in A, and C is a convex set containing A, then any convex combination of the points is again in C. In particular the applies to C = co(A). It’s equally easy to see that if p and q are convex combinations of points in A, then so is any (1 − s)p + sq, where s ∈ [0, 1]. Thus, co(A) can be described intrinsically as the set of all convex combinations of points in A.

1 1.3 Simplexes

N n N An n-simplex in R , denoted σ = σ, is the convex closure of a set of n + 1 points v0, v1, . . . , vn ∈ R in generic position. It will be denoted by ⟨v0, v1, . . . , vn⟩, a notation that will (hopefully) only be used under the proviso that the points are aﬃnely independent.

It’s natural to ask if ⟨v0, v1, . . . , vn⟩ = ⟨w0, w1, . . . , wm⟩ is possible without n = m and w0, w1, . . . wn being some reordering of v0, v1, . . . , vn. The answer is no, as will be demonstrated later on. To wit, a simplex is the convex closure of exactly one set of affinely independent points. This observation makes it clear the vertexes can be used to define properties and derive concepts of simplexes, as is done bellow. ∑ ∑ Thus σ consist of all points p that can be (uniquely) expressed as p = tivi, where ti = 1 and all t1 ∈ [0, 1]. In this connection the scalars ti are called the barycentric coordinates of p and the vi:s are vertexes of σ. The dimension of σ is one less than the number of vertexes, and is thus the same as the dimension of the smallest (affine hyper) plane containing them. A simplex of dimension 0 is hence the same as a single vertex (or point). We allow the set of vertexes to be empty, so that the empty set, ∅, is simplex of dimension −1. The barycenter of σ, or the center of gravity, is the point 1 1 1 1 (v + v + ··· + v ) = v + v + ··· + v ∈ σ. n + 1 0 1 n n + 1 0 n + 1 1 n + 1 n

Any (proper) subset of the vertexes spans a simplex τ which is a (proper) face of σ, written τ ≤ σ (or τ < σ). The boundary of a simplex σ, denoted∑ by ∂σ, is the union of all its proper faces. In terms of convex combinations it consists of all such tivi, where at least one coordinate is 0. Beware that this is generally not the topological boundary of σ as a subset of some RN . ∑ ◦ The interior σ of a simplex σ is σ − ∂σ. This is the set of all tivi, where all ti > 0. This is generally not the same as the interior of σ in a containing Euclidean space.

Lemma 1.1 A simplex of dimension > 0 is (homeomorphic to) the cone on its boundary.

Proof Let σ = ⟨v0, v1, . . . , vn⟩ denote the simplex and b its barycenter. Deﬁne σ v3 f : ∂σ ×I → σ by f(p, t) = (1−t)p+tb. The target is correct by the convexity of σ and f is continuous since it is the restriction of the continuous map t b v2 RN × → RN I , given by the same formula. Obviously, f maps the top of the q v0 cylinder on ∂σ to b. The usual compact-Hausdorﬀ argument gives that, in p ∼ order to prove that f induces a homeomorphism C(∂σ) = σ, we are left with v1 checking that f restricts to a bijection ∂σ × [0, 1) → σ − {b}. Figure 2: Polar coordinates First note that if q ≠ b is a point in σ and q = tb + (1 − t)p, where 0 ≤ t < 1 for q and p ∈ ∂σ, then t/(n + 1) is the minimal barycentric coordinate of q, as some coordinate of p is 0. This shows the uniqueness of t and hence of p, so the restriction of f is injective. On the other hand with q as above, let t¯ < 1/(n + 1) be the minimal barycentric coordinate of q. Write tˆ= (n + 1)t¯ < 1. Then

∑ ∑ ti − t¯ q = (n + 1)tb¯ + (ti − t¯)vi = tbˆ + (1 − tˆ) vi = tbˆ + (1 − tˆ)p. 1 − tˆ i i

Notice that the sum of the barycentric coordinates of p is 1, and that at least one of them is 0, so p is a point in ∂σ. Hence the restriction of f is surjective. If q ∈ σ is q = tp + (1 − t)b, with p ∈ ∂σ and 0 ≤ t ≤ 1, then (t, p) are the polar coordinates for q. They are unique, with the exception that (0, p) corresponds to b for any p ∈ ∂σ The standard n-simplex ∆n in Rn+1 is the convex closure of the standard basis n+1 of R , which in this connection is enumerated e0, e1, . . . , en. Explicitly, this e2 ∆2 e 2 1

Figure 3: The standard 2- simplex sits in R3 means that ∑ n ∆ = {(t0, t1, . . . , tn) | ti = 1 and ti ∈ [0, 1], for all i}.

Lemma 1.2 An n-simplex is homeomorphic to the standard n-simplex hence compact. ∑ n Proof Deﬁne f : ∆ → ⟨v0, v1, . . . , vn⟩ by f(t) = tivi. Then f is continuous. In fact, if v0, v1, . . . vn ∈ Rm the same formula gives a continuous function Rn+1 → Rm. The map f is obviously a bijection, ∆n compact and ⟨v0, v1, . . . , vn⟩ Hausdorﬀ, so f is a homeomorphism.

Lemma 1.3 An n-simplex (n ≥ 1) is homeomorphic to the cone on any of its n − 1-dimensional faces.

Proof Let σ be an n+1-simplex with vertexes v0, v1, . . . , vn and τ the face with the same vertexes except vn. Deﬁne a map f : τ × I → σ by f(p, t) = (1 − t)p + tvn. It’s injective, by uniqueness of barycentric coordinates, with the exception that f(p, 1) = en, for all p. All we need to check to see that f induces a homeomorphism on from the cone on τ to σ is the surjectivity of f| : τ × [0, 1) → σ − {vn}. ∑ If q = tivi ∈ σ and p ≠ vn, i.e. tn ≠ 1, then ∑ ti q = tnvn + (1 − tn) vi = tnvn + (1 − tn)p. 1 − tn i≠ n

Notice that the (positive!) barycentric coordinates of p has sum 1 and at least one is 0. Thus p ∈ τ and f(p, tn) = q.

Lemma 1.4 An n-simplex is homeomorphic to an n-disk by a homeomorphism respecting the ”bound- aries” of the two spaces.

Proof It’s enough to show this for the standard n-simplex ∆n. This simplex is the graph of the map n n n n f : T → R, f(t0, . . . , tn−1) = 1 − Σiti over the set T ⊂ R consisting of those t ∈ R with each n n coordinate ti ∈ [0, 1] and Σiti ≤ 1. Thus ∆ is homeomorphic to T . n n Write |t|s = Σiti and |t|m = max(t0, . . . , tn−1) Deﬁne g : T → I by g(t) = (|t|s/|t|m)t. It’s obviously continuous except possibly at t = 0. But |g(t)|m = |t|s → 0, as t tends to 0. The map g is easily checked −1 n n to be a homeomorphism (with inverse g (p) = (|p|m/|p|s)p). Now, I is homeomorphic to [−1, 1] which in turn is homeomorphic to Dn by the usual normalization. ∼ − Notice that the homeomorphism ∆n → Dn restricts to a homeomorphism δ∆n = δDn = Sn 1. Suppose you have a set A in RN which you some how know to be a simplex. How do you determine it’s vertexes? Are they indeed uniquely 1determined by the set A itself?

Lemma 1.5 A point p in a simplex is a vertex exactly when every line through it intersects the simplex in a subset of the line having p as a boundary point.

Proof Let p be a point of the simplex. The intersection of all faces (with respect to some choice∑ of vertexes) containing p is the minimal face containing p. Suppose this is ⟨v0, v1, . . . , vn⟩ and p = tivi where all ti > 0 by minimality. Suppose t0 < 1, then

∑ t p = t v + (1 − t ) i v = t v + (1 − t )q, 0 0 0 1 − t i 0 0 0 i>0 0 where the sum, call it q, is a convex combination of vertexes and hence in the simplex. Thus the line trough p and q intersects the simplex in a line segment containing v0 and q with p in between.

Thus if p satisﬁes the criterion of the lemma t0 has to be 1 and p = v0.

3 On the other hand let v0 be one of∑ he chosen vertexes and suppose a line intersects the simplex in another point p = tivi. The line through v0 and p is parametrized by ∑ ∑ (1 − s)v0 + s tivi = (1 − s(1 − t0))v0 + stivi, i>0 which is an aﬃne combinations of the vertexes for any real s. In order for this to be a Figure 4: Testing convex combination (and hence a point of the simplex) we need 0 ≤ 1 − s(1 − t0) ≤ 1 for vertex and 0 ≤ sti ≤ 1, for all 1 ≤ i. This implies s ≥ 0, so v0 is a boundary point for the intersecting line segment.

1.4 Simplicial Complexes

A (geometric) simplicial complex in RN is a collection K of simplexes (in RN ) such that

1. if σ is in K, so is any of its faces, 2. if σ and τ are in K, then σ ∩ τ is a face of both σ and τ (which might be the empty simplex).

In this course all simplicial complexes will be ﬁnite (i.e. K is a ﬁnite collection). A subcomplex L of K,L ⊂ K, is a subset closed under taking faces. For example let n be a positive integer and let K(n) be the sub-collection of K consisting of all simplexes in K of dimension ≤ n. Then K(n) is a subcomplex of K, called the n-skeleton of K. The dimension, dimK, of K is the largest dimension of the simplexes in K.

If σ is a simplex of K, let Kσ be the collection of all faces of σ. Then Kσ is a subcomplex of K. It might seem a bit odd to require 1), but it makes it easy to define the notion of ”subcomplex”. A simplex in K is a facet if it is not a proper face of any other simplex of K. The facet obviously determines the rest of K by taking faces. One way to specify a simplicial complex is to pinpoint some simplexes (facets) that pairwise intersect in common faces. Taking all the faces of the facets, to satisfy 1), then defines a simplicial complex. Indeed, 2) follows from the proper intersection of facets, since faces of a fixed simplex always intersect in a common face. The set of zero dimensional simplexes in K is the set of vertexes of K and we denote it by V (K). The polyhedron |K| defined by K, or underlying K, is the union of all simplexes in K : ∪ |K| = σ ⊂ RN . σ∈K Notice that |K| is a subset of RN , as opposed to K which is a collection of subsets |K|. Obviously the polyhedron can equally be described as the union of all facets of K. The polyhedron is compact since it is a finite union of (compact) simplexes. In general a subset of RN is a polyhedron if it is |K| for some simplicial complex K. In this case K is a triangulation of the polyhedron. Different simplicial complexes can very well have the same underlying polyhedron. In this case we say that they are different triangulations of the same space. Intersecting all simplexes containing a point x in the polyhedron of K, we see that there is a unique smallest simplex containing it. It’s called the carrier of x, and we denote it by carr(x) ∈ K. Obviously, x ̸∈ ∂carr(x), so x is an interior point of its carrier. An alternative characterization of carr(x) is that it is the unique simplex with x in its interior. We have a function carr : |K| → K.

1.5 Subdivision

In this section we will denote the barycenter of a simplex σ by ∑ 1 σˆ = v ∈ σ. n + 1 i i

4 A simplicial complex L is a subdivision of K if |L| = |K| and every simplex of L is contained in a simplex of K. This is (sometimes) written L ▹ K. | | | | In this situation, let p be a point in L = K . Then its carrier with∑ respect to L, carrL(p) is contained ⟨ ⟩ in a smallest simplex µ of K. If∑ carrL(p) = u0, . . . , um then p = i siui, for some 0 < si < 1. Each ⟨ ⟩ ≤ ≤ ̸ ui is in µ = v0, . . . vn , so ui = j tijvj, some 0 tij 1, where for each j∑at least∑ one tij = 0, since otherwise carrL(P ) would be cotained in a proper face of µ. This gives p = j( i sitij)vj where each coeﬃcient of the vertexes is > 0. Thus carrL(p) ⊂ µ = carrK (p). A consequence of this is that any simplex in K is a union of simplexes in L. Indeed, take a point p in σ ∈ K. Then carrL(p) ⊂ carrK (p) ⊂ σ. Note that if L ▹ K and K′ is a subcomplex of K, then the set of simplexes in L contained in simplexes of K′ is a subcomplex L′ of L with L′ ▹ K′.

Lemma 1.6 Suppose K′ is a subcomplex of K and L′ a subdivision of K′. Then there is a subdivision L of K containing L′ as a subcomplex.

Put diﬀerently: A subdivision of a subcomplex can be extended to a subdivision of the total complex. Proof By induction on the number of simplexes in K − K′. ′ Let σ be a simplex of lowest dimension in K − K . Then the faces of ∂σ is a subcomplex K∂σ of K ′ ′ corresponding to a certain subcomplex L∂σ of L which is a subdivision of K∂σ. It suﬃces to show that ′ there is a subdivision of Kσ extending L∂σ. ′ For each simplex ν of L∂σ we let νσˆ which is the simplex spanned by the vertexes of ν and the barycenterσ ˆ of σ. The case when ν =∪∅ gives the ⊂ 0-simplexσ ˆ. Obviously νσˆ σ and by polar coordinates σ = ν νσˆ. Also, as a consequence of the essential uniqueness of these coordinates ν1σˆ ∩ ν2σˆ = (ν1 ∩ ν2)ˆσ and ν1 ∩ (ν2σˆ) = ν1 ∩ ν2. { | ∈ ′ } ∪ { | ∈ ′ } Figure 5: Extending a sub- This shows that the collection Lσ = ν ν L∂σ νσˆ ν L∂σ is a ′ division simplicial complex which is a subdivision of Kσ extending L∂σ. ′ ′ ′ Then L ∪ Lσ is a subdivision of K ∪ {σ} extending L . Proceeding by induction we get a subdivision L of K extending L′.

Now, let K be a simplicial complex. If σ0 < σ1 < . . . < σn is a sequence of faces of a simples σn in K, then the barycentersσ ˆ0, σˆ1,... σˆn are aﬃnely ⟨ ⟩ ⊂ independent and hence spans a simplex σˆ0, σˆ1,... σˆn σ. Let sdK be the σ1 collection of all these for all such sequences of faces of simplexes in K. σˆ3 σˆ1 Using the construction in the poof above it follows (by induction) on that sdK σ3 is a simplicial complex which is a subdivision of K. It’s called the barycentric σˆ0 σ subdivision of K. This process can be repeated and we let sdrK denote the 0 r:th barycentric subdivision of K, which we get by subdividing successively r Figure 6: The sequence times. σ0 < σ1 < σ2 determines We will see that the diameter of the simplexes in sdK is smaller than the a 2-simplex in sdK diameter of the simplexes in K in a uniform way. To do this we ﬁrst need a few observations:

Lemma 1.7 The diameter of a simplex is the maximal distance between its vertexes. ∑ Proof. Let σ = ⟨v0, v1, . . . , vn⟩ be an n-simplex and suppose x = tivi ∈ σ. Denote max{d(vi, vj) | 0 ≤ i, j ≤ n} by d.

Then, for any vertex vi0 ∑ ∑ | − | | − | ≤ x vi0 = ti(vi vi0 ) tid = d. ∑ Now, suppose y = sivi ∈ σ. Then, by the above, ∑ ∑ |x − y| = | si(x − vi)| ≤ sid = d.

5 Deﬁnition 1.1 The mesh of a simplicial complex is the maximum of the diameters of its simplexes:

mesh(K) = max{diam(σ) | σ ∈ K}.

Lemma 1.8 For any n-simplex σ the distance from its barycenter σˆ to any other point in σ is ≤ nd/(n + 1), where d is the diameter of σ

Proof Let v0, v1, . . . , vn be the vertexes of σ. then

1 ∑n nd |σˆ − v | = | (v − v )| ≤ i n + 1 j i n + 1 j=0 ∑ as vj − vi = 0, when j = i. Now, suppose y = sivi ∈ σ. Then, ∑ ∑ nd nd |σˆ − y| = | s (ˆσ − v )| ≤ s = . i i i n + 1 n + 1

Lemma 1.9 Suppose σ0 < σ1 < . . . < σn and d is the diameter of σn, which has dimension m, then the diameter of ⟨σˆ0, σˆ1,..., σˆn⟩ is ≤ md/(m + 1)

′ ′ ′ ′ Proof By induction ⟨σˆ0, σˆ1,..., σˆn−1⟩ has diameter ≤ m d /(m + 1), where m is the dimension of σn−1 and d′ is the diameter. Note that m′/(m′ + 1) = 1 − 1/(m′ + 1) ≤ m/(m + 1), since m > m′, and d′ ≤ d, so m′d′/(m′ + 1) ≤ md/(m + 1). Taking the previous lemma in to account the distance between any two vertexes of ⟨σˆ0, σˆ1,..., σˆn⟩ is ≤ md/(m + 1), so the simplex has diameter ≤ md/(m + 1). It now follows that

N mesh(sdK) ≤ mesh(K), N + 1 where N is the maximal dimension of a simplex of K. This is the same as the maximal dimension of a simplex of sdK, so ( ) N r mesh(sdrK) ≤ mesh(K) N + 1 which approaches 0 as r tends to inﬁnity.

2 Simplicial Maps

Let K and L be two simplicial complexes. A map f : |K| → |L| is simplicial if

1. it maps vertexes to vertexes,

2. if ⟨v0, v1, . . . , vn⟩ ∈ K the (the not necessarily diﬀerent) vertexes f(v0), f(v1), . . . , f(vn) are vertexes of a simplex of L 3. when restricted to a simplex f is aﬃne, i.e., ∑ ∑ f( tivi) = tif(vi). i i

It’s usual to denote simplicial maps by f : K → L (i.e. without the decorations for the polyhedrons). Thus a simplicial map is completely determined by its restriction to the set of vertexes. Conversely any map f : V (K) → V (L) satisfying (2) gives a simplicial map f : |K| → |L|, by turning (3) into a deﬁnition.

6 Deﬁnition 2.1 A simplicial approximation to a map f : |K| → |L| is a simplicial map g : |K| → |L|, such that g(x) ∈ carr(f(x)), for each x ∈ |K|

The last condition is equivalent to: any simplex containing f(x) also contains g(x). It’s not true that any map has a simplicial approximation. We will prove, though, that it has after an iterated subdivision of K. ′ ′ ′ ′ If f restricts to a map f| : |K | → |L |, where K and L are subcomplexes of K and L respectively, then ′ ′ ′ ′ ′ for each point x ∈ |K | the carrier of f(x) will be in L . Thus g(x) is in |L | and g| : |K | → |L | is a simplicial approximation to f|.

Theorem 2.1 If g is a simplicial approximation to f, then f ≃ g.

Proof Deﬁne a map F : |K| × I → RN , where RN is a euclidean space containing |L|, by F (x, t) = (1 − t)f(x) + tg(x). We note that the image of this map is in |L| since for any x there is a simplex of L containing both f(x) and g(x).

Deﬁnition 2.2 For a vertex v of a simplicial complex K deﬁne the open star of v as st(v) = {x ∈ |K| | v ∈ carr(x)}

Alternatively, it’s the union of the interiors of all simplexes containing v as a vertex. The complement of st(v) is the union of the simplexes not having v as a vertex, and so is a closed subset of |K| (being the union of a ﬁnite number of compact simplexes). Obviously, the open stars of all vertexes of K forms an open cover of |K|.

Lemma 2.1 Suppose that v0, v1, . . . , vn are vertexes of a simplicial complex K, then ∩n ⟨v0, v1, . . . , vn⟩ ∈ K ⇔ st(vi) ≠ ∅. i=0

Proof Assume that σ = ⟨v0, v1, . . . , vn⟩ ∈ K thenσ ˆ ∈ st(vi) for i = 0, 1, . . . , n. Thus the intersection is non-empty. ∩ ∈ n Assume that x i=0 st(vi), then each vi is a vertex of carr(x), so ⟨v , v , . . . , v ⟩ is a face of carr(x) and thus a simplex of K. Figure 7: Open star of two 0 1 n vertexes Theorem 2.2 Suppose f : |K| → |L| is a map where K and L are simplicial complexes. Then there is an integer r ≥ 0, such that f has a simplicial approximation g : sdrK → L.

Proof The open cover f −1(st(v)), v ∈ V (L) of the compact metric space |K| has a Lebesgue’s number δ > 0. We can choose r so that mesh(sdrK) ≤ δ/2. Then each simplex of K′ = sdrK has diameter less than δ/2. A consequence of this is that for any vertex w of K′ the open star st(w) has diameter less than δ. This, in turns, tells us that there is a vertex v of L such that f(st(w)) ⊂ st(v). Let g(w) be a choice of such a v. ′ Now, suppose that w0, w1, . . . , wn are the vertexes of a simplex of K . Then ∩ ∩ ∅ ̸= f( st(wi)) ⊂ st(g(wi)) i i ′ so g(w0), g(w1), . . . , g(wn) are vertexes of a simplex of L. Hence g deﬁnes a simplicial map K → L. ∩ ∈ Finally, suppose that w0, w1, . . . , wn are the vertexes of the carrier of x. Then x i st(wi). Applying f we get ∩ ∩ ∩ f(x) ∈ f( st(wi)) ⊂ f(st(wi)) ⊂ st(g(wi)) i i i

Hence each vertex g(wi) is in the carrier of f(x). As the carrier of g(x) is the simplex spanned by the ′ g(wi) we get carr(g(x)) ⊂ carr(f(x)), for each x ∈ |K | and g is a simplicial approximation to f.

7 3 Consequences of simplicial approximation

A space X is triangulable if it is homeomorphic to |K| for some simplicial complex K. A specific choice ∼ of such a K and homeomorphism f : X →= |K| gives a triangulation of X. n n For example the n-sphere s is triangulable since it is homeomorphic to ∂∆ = |Kσ|, where Kσ is the simplicial complex of all proper faces of the standard simplex ∆n. It’s a theorem that all smooth manifolds (not defined in this course), which are the fundamental objects of study in differential geometry are triangulable. Thus triangulable spaces exists in abundance.

Theorem 3.1 The set of homotopy classes of maps between two triangulable spaces is countable.

Proof It suﬃces to show that the set of homotopy classes of maps |K| → |L| is countable for any two simplicial complexes K and L. We know that any map |K| → |L| is homotopic to a simplicial map K′ → L, where K′ is some iterated barycentric subdivision of K. Now there is only a ﬁnite number of such map for any K′, but a countable collection of iterated barycentric subdivisions of K.

Theorem 3.2 The inclusion of the 2-skeleton i : |K(2)| ,→ |K| induces an isomorphism on fundamental groups for any simplicial complex K.

Proof The space I is the polyhedron of a simplicial complex with two vertexes and one 1-simplex. Any path-component of |K| contains a vertex v, so we can consider loops at this point in |K|. Any such loop ω : I → |K| then maps the subcomplex consisting of 0 and 1 to the subcomplex of K consisting of just the vertex v. Iterated barycentric subdivision of I means introduction of (equidistant) points in I. Thus ω is homotopic to a map ω′ : I → |K| which is simplicial. As ω maps the subcomplex consisting of 0 and 1 to the subcomplex consisting of v the so will ω′. The usual homotopy between ω and ω′ is therefore stationary on the endpoints of I. Thus ω and ω′ are in fact path-homotopic. ′ (1) (2) As the image of ω is contained in |K |, this shows that i induces a surjection i∗ : π1(|K |, v) → π1(|K|, v). 2 To show injectivity, consider I as the polyhedron of the simplicial complex having the vertexes v0 = (0, 0), v1 = (1, 0), v2 = (1, 1) and v3 = (0, 1), the 1-simplexes ⟨v0, v1⟩, ⟨v1, v2⟩, ⟨v1, v2⟩ and ⟨v2, v3⟩, and the 2-simplexes ⟨v0, v1, v2⟩ and ⟨v0, v2, v3⟩. It has a subcomplex J consisting of the faces of the 1-simplexes ⟨v0, v1⟩, ⟨v1, v2⟩, ⟨v2, v3⟩, and ⟨v1, v3⟩. Now suppose that α, β : I → |K(2)| are loops at the vertex v, which are path-homotopic in |K|. Let F : I2 → |K| be a path-homotopy between α and β in |K|. It’s restriction to J has image in |K(2)| and its restriction to J ′ is constant (with value v). Thus if F ′ : I2 → |K| is a simplicial approximation to F after some subdivision of I2, the same is true for F ′. Even more importantly, the image of F ′ has image in |K(2)|, since it’s simplicial. From this is follows that α′(t) = F ′(t, 0) and β′(t) = F ′(t, 1) are a simplicial approximation to α and β in |K(2)|. The new paths α′ and β′ are in fact path-homotopic (in |K(2)|) to α and β, respectively. As F ′ is a path homotopy α′ ≃ β′ (in |K(2)|) we get (in |K(2)|)

α ≃ α′ ≃ β′ ≃ β,

so i∗ is injective.

Corollary 3.1 The n-sphere Sn is simply connected if n ≥ 2.

Proof Sn is homeomorphic to the polyhedron ∂∆n. When n ≥ 2 ∂∆n and ∂∆n has the same 2-skeletons. ∼ Hence the fundamental group of Sn is isomorphic to the fundamental group of ∆n = Dn, which is trivial since Dn is homotopy equivalent to a point.