Homology

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Simplices

First, we introduce simplices, which provide intuitive guidance when developing homology theory.

Definition 1 For any natural number \(k\), suppose \(k+1\) points \(v_0,\ldots, v_k\in\mathbb{R}^d\) in general position are given. Then the \(k\)-simplex formed by these points is the smallest convex set containing \(\{v_0,\ldots, v_k\}\).

Here, saying that the \(k+1\) vertices are in general position means that they do not lie in any hyperplane of dimension less than \(k\). Equivalently, the \(k\) vectors

\[v_1-v_0,\ldots, v_k-v_0\]

are linearly independent. For example, a \(0\)-simplex is a point, a \(1\)-simplex is a line segment joining two vertices, a \(2\)-simplex is a triangle joining three points, and a \(3\)-simplex is a tetrahedron.

As shown in the figure, the \(n\)-simplex in \(\mathbb{R}^{n+1}\) formed by the \(n+1\) vertices

\[(1,0,\ldots, 0),\qquad\cdots,\qquad (0,0,\ldots,1)\]

is called the standard simplex. Naturally, these simplices themselves are not our primary objects of interest; rather, we aim to use them to compute invariants of manifolds. To this end, we must define a \(\Delta\)-complex structure on a manifold, and we begin with a simple, intuitive example.

Example 2 A common example is the (2-dimensional) torus \(T^2\). Its elementary definition is the product manifold \(S^1\times S^1\), but to see intuitively that this product is a torus, consider the following figure.

In this figure, imagine gluing the edges of each color together in the direction of the arrows “without twisting.” For instance, first glue the horizontal edges to form a cylinder, then glue the remaining edges along the top and bottom of the cylinder to obtain the following.

If the above square were a subset of the coordinate plane passing through the four points \((0,0),(0,1),(1,0),(1,1)\), then this is the quotient space obtained by imposing the equivalence relation

\[(x,0)\sim (x,1),\qquad (0,y)\sim(1,y)\]

on the square. Thinking of \(S^1\) as the quotient of the interval \([0,1]\) on the line obtained by identifying \(0\) and \(1\), we see that this agrees with the definition \(T^2=S^1\times S^1\) above. On the other hand, if on the same figure we reverse one edge and glue only the horizontal edges, obtaining the following quotient space,

the resulting space is not a cylinder but a Möbius strip

In the above example, dividing the squares drawn on the plane into two triangles along a diagonal allows us to view the squares as two \(2\)-simplices glued together; transferring this to the quotient space in the same manner lets us understand the spaces in Example 2 as built by gluing simplices. As this example shows, when gluing \(2\)-simplices the direction of the edges matters—more generally, when gluing \(n\)-simplices, the direction of the \((n-1)\)-simplex matters—and this is determined by imposing a total order on the vertices. For instance, if we declare the \(k\)-simplex formed by vertices \(v_0,\ldots, v_k\) to be positively oriented when listed in index order, then applying an odd permutation to obtain \(v_1,v_0,v_2,\ldots,v_k\) yields the negative orientation. We write a simplex with orientation given by the index order as \([v_0,\ldots, v_k]\). Under this notation, whether we view the \((k-1)\)-simplex obtained by forgetting one vertex of the \(k\)-simplex as a face of the original \(k\)-simplex or as the following \((k-1)\)-simplex

\[[v_0,\ldots,\hat{v}_i,\ldots, v_k]=[v_0,\ldots, v_{i-1},v_{i+1},\ldots, v_k]\]

its orientation is the same.

Definition 3 For a topological space \(X\), a \(\Delta\)-complex structure on it is a collection of maps \(\sigma_\alpha:\Delta^{n(\alpha)}\rightarrow X\) satisfying the following.

1. The restriction of \(\sigma_\alpha\) to \(\interior(\Delta^{n(\alpha)})\) is injective, and for every point \(x\) in \(X\) there exists exactly one \(\alpha\) such that \(x\in \sigma_\alpha(\interior(\Delta^{n(\alpha)}))\).
2. The restriction of \(\sigma_\alpha\) to a face of \(\Delta^{n(\alpha)}\), namely \(\sigma_\alpha\vert_{\Delta^{n(\alpha)-1}}:\Delta^{n(\alpha)-1}\rightarrow X\), also belongs to this collection.
3. A subset \(A\subseteq X\) is open in \(X\) if and only if \(\sigma_\alpha^{-1}(A)\) is open in \(\Delta^{n(\alpha)}\) for every \(\alpha\).

For example, the standard simplex \(\Delta^2\) trivially carries a \(\Delta\)-complex structure; explicitly, the maps giving this structure are

\[\operatorname{id}_{\Delta^2}:\Delta^2\rightarrow\Delta^2\]

together with three maps \(\sigma^1_1,\sigma^1_2,\sigma^1_3\) sending the \(1\)-simplex \(\Delta^1\) to each edge, and three maps \(\sigma_1^0,\sigma_2^0,\sigma_3^0\) sending the \(0\)-simplex \(\Delta^0\) to each vertex.

Example 4 For instance, the 2-dimensional torus \(T^2\) can be represented as in the following figure,

and this figure simultaneously gives a \(\Delta\)-complex structure on \(T^2\).

Simplicial Homology

We now define invariants of topological spaces using the \(\Delta\)-complex structure introduced above. More concretely, these will be defined through groups of formal sums of simplices. However, a subtle issue arises here: for this to be an invariant of the topological space \(X\), it must not depend on the choice of \(\Delta\)-complex structure.

For instance, if we subdivide the square in Example 2 more finely to create additional 2-simplices as in the figure above, the new gray \(1\)-simplices that appear must cancel each other out in some fashion. That is, when choosing a \(\Delta\)-complex we glue them with appropriately matched orientations, but when computing the invariant we must add them with opposite signs. Keeping this in mind will make the following computation more transparent.

Consider a topological space \(X\) with a \(\Delta\)-complex structure, and let \(C^\Delta_k(X)\) be the free abelian group generated by the \(k\)-simplices. That is,

\[C^\Delta_k(X)=\{\sigma_\alpha:\Delta^{n(\alpha)}\rightarrow X\text{ $k$-simplex}\mid n(\alpha)=k\}\cdot\mathbb{Z}\]

Following the convention for abelian groups, the operation on \(C^\Delta_k(X)\) is written as addition. As we saw earlier, the boundary of a \(k\)-simplex \([v_0,\ldots, v_k]\) consists of the following \((k-1)\)-simplices:

\[[v_1,v_2,\ldots, v_k],\quad[v_0,v_2,\ldots, v_k],\quad\cdots,[v_0,v_1,\ldots\hat{v}_i,\ldots,v_k],\quad\cdots,\quad[v_0,v_1,\ldots, v_{k-1}]\]

If we think of the boundary of \([v_0,\ldots, v_k]\) as merely the sum of these, we obtain a function from \(C^\Delta_k(X)\) to \(\Delta_{k-1}(X)\). However, if we define the boundary map \(\partial_k\) not as a simple sum but by the formula

\[\partial_k(\sigma_\alpha\vert_{[v_0,\ldots,v_k]})=\sum_{i=0}^n(-1)^i\sigma_\alpha\vert_{[v_0,\ldots, \hat{v}_i,\ldots,v_k]}\tag{1}\]

then it is well known that \((C^\Delta_k(X),\partial_k)\) forms a chain complex. The signs on the right-hand side of this formula are arranged so that adjacent simplices inside \(X\) cancel each other, as noted above.

Proposition 5 \((C^\Delta_k(X),\partial_k)\) is a chain complex. (§Abelian Categories, ⁋Definition 4)

Proof

For any \(\sigma_\alpha\in C^\Delta_k(X)\), since

\[\partial_k(\sigma_\alpha\vert_{[v_0,\ldots,v_k]})=\sum_{i=0}^n(-1)^i\sigma_\alpha\vert_{[v_0,\ldots, \hat{v}_i,\ldots,v_k]}\]

we have

\[\partial_{k-1}\partial_k(\sigma_\alpha\vert_{[v_0,\ldots,v_k]})=\sum_{j < i}(-1)^{i+j}\sigma_\alpha\vert_{[v_0,\ldots, \hat{v}_j,\ldots\hat{v}_i,\ldots,v_k]}+\sum_{j > i}(-1)^{i+j-1}\sigma_\alpha\vert_{[v_0,\ldots, \hat{v}_i,\ldots\hat{v}_j,\ldots,v_k]}\]

and thus the first and second sums cancel each other and vanish.

The \(n\)th homology of the chain complex \((C^\Delta_k(X),\partial_k)\) obtained in this way is called the \(n\)th simplicial homology, and is denoted \(H_n^\Delta(X)\). (§Homology, ⁋Definition 2)

Example 6 Represent the 2-dimensional torus \(T^2\) as in Example 4 above, and let the \(2\)-simplices in the order listed on the right be \(L\) and \(U\), the \(1\)-simplices be \(a,b,c\), and the \(0\)-simplex be \(p\). Since directions are already assigned to the two \(1\)-simplices \(b,c\), for these to be simplices \(a\) must be directed from lower left to upper right. Now, assuming the vertices of \(U\) are given as \(v_0,v_1,v_2\) as in the figure

we may set

\[a=[v_0,v_2],\quad b=[v_1,v_2],\quad c=[v_0,v_1]\]

Similarly, assuming the vertices of \(L\) are given as \(v_0,v_1,v_2\) as in the following figure

for \(L\) we may take

\[a=[v_0,v_2],\quad b=[v_0,v_1],\quad c=[v_1,v_2]\]

Now considering the boundary map \(\partial_2:C^\Delta_2(T^2)\rightarrow C^\Delta_1(T^2)\), we have

\[\begin{aligned}\partial_2(U)&=[v_1,v_2]-[v_0,v_2]+[v_0,v_1]=b-a+c,\\ \partial_2(L)&=[v_1,v_2]-[v_0,v_2]+[v_0,v_1]=c-a+b\end{aligned}\]

Also, considering \(\partial_1:C^\Delta_1(T^2)\rightarrow C^\Delta_0(T^2)\), since all these vertices correspond to the same point \(p\) in \(T^2\),

\[\partial_1(a)=\partial_1(b)=\partial_1(c)=p-p=0\]

and thus in the complex

\[\cdots\overset{\partial_3}{\longrightarrow}C^\Delta_2(T^2)=\langle L,U\rangle\overset{\partial_2}{\longrightarrow}C^\Delta_1(T^2)=\langle a,b,c\rangle\overset{\partial_1}{\longrightarrow}C^\Delta_0(T^2)=\langle p\rangle\overset{\partial_0}{\longrightarrow}0\]

we have

\[\ker\partial_2=\langle L-U\rangle,\qquad\ker\partial_1=C^\Delta_1(T^2),\qquad\ker\partial_0=C^\Delta_0(T^2)\]

and

\[\im\partial_3=0,\qquad\im\partial_2=\langle a-b-c\rangle,\qquad \im\partial_1=0\]

so

\[H_2^\Delta(T^2)=\ker\partial_2/\im\partial_3\cong\mathbb{Z},\quad H_1^\Delta(T^2)=\ker\partial_1/\im\partial_2\cong \mathbb{Z}\oplus\mathbb{Z},\quad H_0^\Delta(T^2)=\ker\partial_0/\im\partial_1\cong\mathbb{Z}\]

and the remaining homology groups are all \(0\).

Meanwhile, in §Homology, ⁋Definition 2, we called elements of \(Z_n(C)=\ker\partial_n\) \(n\)-cycles and elements of \(B_n(C)=\im\partial_{n+1}\) \(n\)-boundaries, and now their names are intuitively clear. Namely, here the boundary maps actually compute the boundary faces of simplices; \(n\)-cycles are those whose values cancel upon computing these boundary faces—for instance, in Example 6, \(a,b\) (and \(c\)) represent closed curves in the original space \(T^2\)—and \(n\)-boundaries literally mean \(n\)-simplices that appear as the boundary of some \((n+1)\)-simplex.

Singular Homology

The simplicial homology defined above has an intuitively clear meaning, but it suffers from an evident limitation: to compute the homology of an arbitrary topological space \(X\), we must equip it with a \(\Delta\)-complex structure. Even if \(X\) is a topological manifold, it is well known that when its dimension is \(4\) or higher, giving a \(\Delta\)-complex structure on \(X\) may be impossible.

Therefore, we relax this condition to define a new homology.

Definition 7 A singular \(k\)-simplex on \(X\) is a continuous map \(\sigma:\Delta^k\rightarrow X\).

Unlike the \(\Delta\)-complex structure defined earlier, singular \(k\)-simplices need not maintain the shape of a \(k\)-simplex inside \(X\) at all. For example, a constant map sending all points of \(\Delta^k\) to a single point is also a singular \(k\)-simplex.

Now let \(C_k(X)\) be the free abelian group generated by all singular \(k\)-simplices, and define \(\partial_k:C_k(X)\rightarrow C_{k-1}(X)\) in the same manner as in formula (1) above. Then we can verify in exactly the same way as in Proposition 5 that \((C_k(X), \partial_k)\) forms a chain complex, and the resulting homologies are called singular homology and denoted \(H_n(X)\).

Example 8 Computing singular homology directly from the definition is not advisable, but let us carry out a bit of (non-rigorous) computation for intuition.

By definition, a singular \(0\)-simplex on an arbitrary topological space \(X\) is a continuous map from \(\Delta^0\) to \(X\). Since \(\Delta^0\) is just a single point, \(C_0(X)\) is the free abelian group generated by the points of \(X\). Similarly, identifying \(\Delta^1\) with the interval \(I=[0,1]\), we see that \(C_1(X)\) is simply the free abelian group generated by paths in \(X\); here we also allow constant paths \([0,1]\rightarrow X\), and it is for this reason that we call these singular \(1\)-simplices. Likewise, after continuous deformation, \(C_2(X)\) becomes the free abelian group generated by disks contained in \(X\).

Then the boundary of a path \(\sigma:[0,1]\rightarrow X\) is given by \(\partial_1\sigma=\sigma(1)-\sigma(0)\) according to formula (1) above, whence

\[Z_1(X)=\ker\partial_1=\left\{\sigma:[0,1]\rightarrow X\mid \sigma(1)=\sigma(0)\right\}\]

That is, intuitively, \(Z_1(X)\) can be thought of as the subgroup generated by closed curves in \(X\). Similarly, if we assign a geometric meaning to \(B_1(X)=\im\partial_2\), it consists of closed curves in \(X\) that bound some disk; thus the first homology

\[H_1(X)=\frac{Z_1(X)}{B_1(X)}\]

measures how many closed curves in \(X\) do not bound disks. For example, the first homology of the subset of \(\mathbb{R}^2\)

\[D^2=\left\{(x,y)\in \mathbb{R}^2\mid x^2+y^2\leq 1\right\}\]

is \(0\), because any closed curve in \(D^2\) can trivially be filled in.

On the other hand, the first homology of the space \(D^2\setminus {(0,0)}\) is not \(0\). For instance, considering the following closed curve

there is no way to fill its interior continuously to make it a disk. However, if we similarly consider the punctured space

\[D^3\setminus \left\{(0,0,0)\right\}=\left\{(x,y,z)\in \mathbb{R}^3\mid 0< x^2+y^2+z^2\leq 1\right\}\]

the first homology of this space is \(0\), because even when given a closed curve “containing the hole”

\[S^1=\left\{(x,y,0)\in \mathbb{R}^3\mid x^2+y^2=1\right\}\]

as in the figure

we can view it as the boundary of a disk. Instead, the second homology of this space will not be \(0\).

There is a minor gap in this computation. For example, a constant map sending all points of \(\Delta^1\) to a fixed \(x\in X\) is a singular \(1\)-simplex by definition, yet when we claim that any closed curve in \(D^2\) can be filled, we did not consider (constant) paths of this form. However, what we ultimately need to show is that applying \(\partial\) to a suitable singular \(2\)-simplex \(\Delta^2 \rightarrow X\) yields this singular \(1\)-simplex; thus we simply consider a \(2\)-simplex sending all points of \(\Delta^2\) to the fixed \(x\in X\), and then take its boundary to obtain exactly the desired complex \(\Delta^1 \rightarrow X\). Generalizing this, for arbitrary \(n\), a singular \(n\)-simplex sending all points of \(\Delta^n\) to a fixed \(x\in X\) is the boundary of a singular \((n+1)\)-simplex sending all points of \(\Delta^{n+1}\) to the fixed \(x\in X\). In other words, a constant map always becomes the identity in \(H_n(X)\).

Gaps of this kind can be resolved with a little care as above. In fact, for any space \(X\) admitting a \(\Delta\)-complex structure, singular homology \(H_n(X)\) and simplicial homology \(H_n^\Delta(X)\) always coincide. Roughly speaking, this is because singular maps \(\Delta^k \rightarrow X\) such as constant maps become boundaries of similarly singular \(\Delta^{k+1}\rightarrow X\), so that when comparing the two quotients

\[H_n^\Delta(X)=\frac{\ker\partial_n^\Delta}{\im\partial_{n+1}^\Delta},\qquad H_n(X)=\frac{\ker\partial_n}{\im\partial_{n+1}}\]

allowing singular \(\Delta^k \rightarrow X\) enlarges \(\ker \partial_n\) over \(\ker\partial_n^\Delta\) by exactly as much as \(\im\partial_n\) also enlarges, so the two quotients end up equal.

A more fundamental issue is that this computation relies entirely on our geometric intuition, and to compute the homology of more complicated spaces we must study more general properties of homology.

Properties of Homology

Proposition 9 When a topological space \(X\) is expressed as a disjoint union of path-components \(X=\coprod X_i\), the following isomorphism holds:

\[H_n(X)\cong \bigoplus_{i\in I} H_n(X_i)\]

Proof

First, since the continuous image of a path-connected space \(\Delta^k\) is path-connected, the images of singular simplices lie entirely within the \(X_i\). Hence \(C_n(X)\cong \bigoplus_{i\in I} C_n(X_i)\). For the same reason, the \(\partial\) maps preserve this decomposition, and since direct sums preserve the kernels and images of such maps, we obtain the desired result.

Thus computing the homology of an arbitrary topological space reduces to computing the homology of a path-connected space. However, this is still not an easy problem. We cannot compute it in general, but the case \(n=0\) has geometric meaning.

Proposition 10 For a non-empty path-connected space \(X\), we have \(H_0(X)\cong \mathbb{Z}\).

Proof

First, since \(\partial_0=0\),

\[H_0(X)=\ker\partial_0/\im\partial_1=C_0/\im\partial_1\]

To construct an isomorphism \(H_0(X)\rightarrow\mathbb{Z}\), define a homomorphism \(\varepsilon:C_0(X)\rightarrow\mathbb{Z}\) by

\[\varepsilon\left(\sum n_i\sigma_i\right)=\sum_i n_i\]

Since \(X\) is non-empty, \(\varepsilon\) is surjective. Therefore, by the first isomorphism theorem, it suffices to show that \(\ker\varepsilon=\im\partial_1\). That \(\ker\varepsilon\) contains \(\im\partial_1\) is immediate from the definition of \(\partial_1\), so we need only show the reverse inclusion. Assume \(\varepsilon\left(\sum n_i\sigma_i\right)=0\), and let \(x_i\) be the image of each \(0\)-simplex \(\sigma_i\). Then by path-connectedness of \(X\), we can choose a point \(x\) and paths from \(x\) to each \(x_i\); these determine \(1\)-simplices in the corresponding direction. Denoting these by \(\tau_i\), we have \(\partial \tau_i=\sigma_i-\sigma\), and thus

\[\partial\left(\sum_i n_i\tau_i\right)=\sum n_i\sigma_i-\sum n_i \sigma=0\]

which yields the desired result.

The proof is somewhat lengthy for the sake of rigor, but the essential idea is simple: in a path-connected space \(X\), any two points can be joined by a path, and viewing this path as a \(1\)-simplex, the two points become the boundary of that \(1\)-simplex; hence for \(B_0(X)=\im\partial_1\) we may regard these two points as equal.

Conversely, there are cases where we can compute the homology for all \(n\), namely when \(X\) is a single point. In this case, regardless of \(k\), a singular \(k\)-simplex \(\sigma_k:\Delta^k \rightarrow X\) is uniquely determined (namely, as a constant map), and considering formula (1), \(\partial_k\) is \(0\) when \(k\) is odd and sends \(\sigma_k\) to \(\sigma_{k-1}\) when \(k\) is even. Thus the chain complex of singular simplices is

\[\cdots\rightarrow \mathbb{Z} \overset{0}{\longrightarrow}\mathbb{Z}\overset{\approx}{\longrightarrow} \mathbb{Z}\overset{0}{\longrightarrow}\mathbb{Z}\rightarrow0\]

and we obtain the following.

Proposition 11 For a one-point space \(X\), we have \(H_0(X)\cong \mathbb{Z}\) and \(H_k(X)\cong 0\) for all \(k>0\).

Yet the most important property, arguably, is functoriality. We already know that for each \(n\), computing the \(n\)th homology \(H_n:\Ch_{\geq 0}(\Ab)\rightarrow \Ab\) is a functor on the category of chain complexes of abelian groups \(\Ch_{\geq 0}(\Ab)\). Therefore, to show that the composition

\[\Top \rightarrow \Ch_{\geq 0}(\Ab)\rightarrow \Ab\]

is a functor, it suffices to show that \(\Top \rightarrow \Ch_{\geq 0}(\Ab)\) is a functor.

Proposition 12 \(\Top\rightarrow\Ch_{\geq 0}(\Ab)\) is a functor.

Proof

That is, we must show that for any continuous map \(f:X\rightarrow Y\), there exists a chain map \(C_\bullet(f):C_\bullet(X)\rightarrow C_\bullet(Y)\). Naturally, we define \(C_\bullet(f)\) by

\[C_\bullet(f):\sigma\mapsto f\circ\sigma\]

and the key is to verify that this is a chain map. This follows simply because for any \(\sigma:\Delta^n \rightarrow X\),

\[\begin{aligned}(C_\bullet(f)\circ\partial^X_n)(\sigma)&=C_\bullet(f)\left(\sum_{i=0}^n(-1)^i\sigma\vert_{[v_0,\ldots,\hat{v}_i,\ldots,v_n]}\right)=\sum_{i=0}^n(-1)^iC_\bullet(f)(\sigma\vert_{[v_0,\ldots,\hat{v}_i,\ldots,v_n]})\\&=\sum_{i=0}^n(-1)^i f\circ(\sigma\vert_{[v_0,\ldots,\hat{v}_i,\ldots,v_n]})=\sum_{i=0}^n(-1)^i (f\circ\sigma)\vert_{[v_0,\ldots,\hat{v}_i,\ldots,v_n]}\\&=\partial_n^Y(f\circ\sigma)=(\partial_n^Y\circ C_\bullet(f))(\sigma)\end{aligned}\]

Finally, we make the following definition.

Definition 13 If \(H_n(X)\cong H_n(Y)\) holds for all \(n\), we say that two spaces \(X,Y\) are homologous.

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References

[Hat] A. Hatcher, Algebraic Topology. Cambridge University Press, 2022.

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