This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Simplices
First, we introduce simplices, which help provide intuitive understanding when developing homology theory.
Definition 1 For any natural number \(k\), suppose we are given \(k+1\) points \(v_0,\ldots, v_k\in\mathbb{R}^d\) arranged in general position. Then the \(k\)-simplex formed by these points is the smallest convex set containing the set \(\{v_0,\ldots, v_k\}\).
Here, saying that the \(k+1\) vertices are arranged in general position means that these points do not lie in any hyperplane of dimension less than \(k\). Equivalently, this can be understood as the \(k\) vectors
\[v_1-v_0,\ldots, v_k-v_0\]being linearly independent. For example, a \(0\)-simplex is a point, a \(1\)-simplex is a line segment connecting two vertices, a \(2\)-simplex is a triangle connecting three points, and a \(3\)-simplex is a tetrahedron.
As shown in the figure above, the \(n\)-simplex formed by \(n+1\) vertices
\[(1,0,\ldots, 0),\qquad\cdots,\qquad (0,0,\ldots,1)\]in \(\mathbb{R}^{n+1}\) is called the standard simplex. Naturally, these simplices themselves are not our primary objects of interest; rather, we are interested in using them to compute invariants of manifolds. To this end, we need to define a \(\Delta\)-complex structure on a manifold, and we begin with a simple and intuitive example.
Example 2 A common example we often consider is the (2-dimensional) torus \(T^2\). Its simple definition is the product manifold \(S^1\times S^1\), but to see intuitively that this product manifold is a torus, we can think of the following figure.
In this figure, imagine gluing the edges of each color together in the direction of the arrows “without twisting.” For instance, if we first glue the horizontal edges to form a cylinder and then glue the remaining edges along the top and bottom of the cylinder, we obtain the following.
If the above square were points on a 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 obtained by giving the quotient topology on the interval \([0,1]\) on the real line 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 we consider the following quotient space on the same figure, where one edge is reversed and only the horizontal edges are glued,
then the resulting space is not a cylinder but a Möbius strip
In the above example, if we divide the squares drawn on the plane into two triangles along a diagonal, these squares can be thought of as two \(2\)-simplices glued together, and transferring this to the quotient space in the same way as above allows us to understand the spaces in Example 2 as built by gluing simplices together. As seen in this example, when gluing \(2\)-simplices, the direction of the edges (more generally, when gluing \(n\)-simplices, the direction of the \((n-1)\)-simplex) is important, and this is determined by giving a total order among the vertices. For instance, if we say that the \(k\)-simplex formed by vertices \(v_0,\ldots, v_k\) is positively oriented when listed in index order, then applying an odd permutation to obtain \(v_1,v_0,v_2,\ldots,v_k\) gives the negative orientation. We denote a simplex with orientation given according to index order by \([v_0,\ldots, v_k]\). 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 remains the same.
Definition 3 For a topological space \(X\), a \(\Delta\)-complex structure on it is a collection of functions \(\sigma_\alpha:\Delta^{n(\alpha)}\rightarrow X\) defined as follows.
- The restriction of \(\sigma_\alpha\) to \(\interior(\Delta^{n(\alpha)})\) is injective, and for any point \(x\) in \(X\), there exists exactly one \(\alpha\) such that \(x\in \sigma_\alpha(\interior(\Delta^{n(\alpha)}))\).
- 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 of functions.
- A subset \(A\subseteq X\) is open in \(X\) if and only if \(\sigma_\alpha^{-1}(A)\) is open in \(\Delta^{n(\alpha)}\) for each \(\alpha\).
For example, the standard simplex \(\Delta^2\) trivially has a \(\Delta\)-complex structure, and the functions explicitly giving this structure are
\[\operatorname{id}_{\Delta^2}:\Delta^2\rightarrow\Delta^2\]together with three functions \(\sigma^1_1,\sigma^1_2,\sigma^1_3\) sending the \(1\)-simplex \(\Delta^1\) to each edge, and three functions \(\sigma_1^0,\sigma_2^0,\sigma_3^0\) sending the \(0\)-simplex \(\Delta^0\) to each vertex.
Example 4 For example, 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
Now we define invariants of topological spaces using the \(\Delta\)-complex structure defined above. More concretely, these will be defined through groups formed by formal sums of simplices. However, there is a subtle issue to point out 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 more 2-simplices as in the figure above, the new gray \(1\)-simplices that arise must cancel each other out in some way. That is, when choosing a \(\Delta\)-complex, we must glue them with orientations matched appropriately, but when computing the invariant, we must add them with opposite signs. Keeping this in mind will make the following computation a bit more intuitive.
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 when dealing with abelian groups, we consider the operation on \(C^\Delta_k(X)\) to be given by 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 simply 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 of these simplices but by the formula
\[\partial_k(\sigma_\alpha|_{[v_0,\ldots,v_k]})=\sum_{i=0}^n(-1)^i\sigma_\alpha|_{[v_0,\ldots, \hat{v}_i,\ldots,v_k]}\tag{1}\]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 set so that adjacent simplices inside \(X\) cancel each other out, as pointed out above.
Proposition 5 \((C^\Delta_k(X),\partial_k)\) is a chain complex. ([Category Theory] §Abelian Categories, ⁋Definition 4)
Proof
For any \(\sigma_\alpha\in C^\Delta_k(X)\), since
\[\partial_k(\sigma_\alpha|_{[v_0,\ldots,v_k]})=\sum_{i=0}^n(-1)^i\sigma_\alpha|_{[v_0,\ldots, \hat{v}_i,\ldots,v_k]}\]we have
\[\partial_{k-1}\partial_k(\sigma_\alpha|_{[v_0,\ldots,v_k]})=\sum_{j < i}(-1)^{i+j}\sigma_\alpha|_{[v_0,\ldots, \hat{v}_j,\ldots\hat{v}_i,\ldots,v_k]}+\sum_{j > i}(-1)^{i+j-1}\sigma_\alpha|_{[v_0,\ldots, \hat{v}_i,\ldots\hat{v}_j,\ldots,v_k]}\]and thus the first and second sums cancel each other out 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 by \(H_n^\Delta(X)\). ([Homological Algebra] §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 given for the two \(1\)-simplices \(b,c\), for these to be simplices, \(a\) must have the direction 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 can 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 can think of
\[a=[v_0,v_2],\quad b=[v_0,v_1],\quad c=[v_1,v_2]\]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 [Homological Algebra] §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. That is, in this case the boundary maps actually compute the boundary faces of simplices, \(n\)-cycles are those whose values cancel out when the boundary faces are computed in this way—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 has a clear limitation in that to compute the homology of an arbitrary topological space \(X\), we must give a \(\Delta\)-complex structure on it. Even if \(X\) is a topological manifold, it is well known that if its dimension is \(4\) or higher, it may be impossible to give a \(\Delta\)-complex structure on \(X\).
Therefore, we relax this condition to define a new homology.
Definition 7 A singular \(k\)-simplex defined on \(X\) is a continuous function \(\sigma:\Delta^k\rightarrow X\).
Unlike the \(\Delta\)-complex structure defined in the previous section, singular \(k\)-simplices do not need to 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 way 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 homologies in this case are called singular homology and denoted by \(H_n(X)\).
Example 8 Computing singular homology directly from the definition is not a good idea, but let us do a bit of (non-rigorous) computation for our intuition.
By definition, a singular \(0\)-simplex on an arbitrary topological space \(X\) is a continuous function 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 \(I=[0,1]\), \(C_1(X)\) is just the free abelian group generated by paths in \(X\), and in this case we also allow constant paths \([0,1]\rightarrow X\), which is why we call this a singular \(1\)-simplex. 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, and from this we know that
\[Z_1(X)=\ker\partial_1=\left\{\sigma:[0,1]\rightarrow X\mid \sigma(1)=\sigma(0)\right\}\]That is, intuitively, \(Z_1(X)\) can be thought of as the subgroup generated by closed curves in \(X\). Similarly, if we give a geometric meaning to \(B_1(X)=\im\partial_2\), this means closed curves in \(X\) that arise as the boundary of some disk, and thus the first homology
\[H_1(X)=\frac{Z_1(X)}{B_1(X)}\]examines how many closed curves in \(X\) do not arise as boundaries of 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 \(0\). This is because for any closed curve in \(D^2\), there is trivially a way to fill its interior.
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 continuously fill its interior to make it a disk. However, similarly considering the following 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 if we are 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 \(1\)-complex by definition, but 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 need to show is ultimately that applying \(\partial\) to a suitable singular \(2\)-complex \(\Delta^2 \rightarrow X\) yields this singular \(1\)-complex, so if we simply consider a \(2\)-complex sending all points of \(\Delta^2\) to a fixed \(x\in X\) and then take the boundary of this complex, it becomes exactly the complex \(\Delta^1 \rightarrow X\) that we want. Generalizing this, for arbitrary \(n\), a singular \(n\)-complex sending all points of \(\Delta^n\) to a fixed \(x\in X\) is the boundary of a singular \((n+1)\)-complex sending all points of \(\Delta^{n+1}\) to a 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 bit of care as above. In fact, it can be shown that for any space \(X\) admitting a \(\Delta\)-complex structure, singular homology \(H_n(X)\) and simplicial homology \(H_n^\Delta(X)\) always coincide, which roughly speaking 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 \(\Delta^k \rightarrow X\) makes \(\ker \partial_n\) larger than \(\ker\partial_n^\Delta\) by exactly as much as \(\im\partial_n\) also becomes larger, so that the two quotients end up being equal.
A somewhat more fundamental issue is that this computation relies entirely on our geometric intuition, and to compute the homology of more complicated spaces we need to study more general properties of homology.
Properties of Homology
Proposition 9 When a topological space \(X\) is expressed as the 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 \(X_i\). From this we know that \(C_n(X)\cong \bigoplus_{i\in I} C_n(X_i)\). For the same reason, the \(\partial\) maps also 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 an arbitrary 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 \(X\), \(H_0(X)\cong \mathbb{Z}\).
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 the formula
\[\varepsilon\left(\sum n_i\sigma_i\right)=\sum_i n_i\]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 trivial from the definition of \(\partial_1\), so it suffices to 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 the assumption that \(X\) is path-connected, we can choose a suitable point \(x\) and paths from \(x\) to each \(x_i\), and these determine a \(1\)-simplex in this direction. Let these be \(\tau_i\); then \(\partial \tau_i=\sigma_i-\sigma\), and thus
\[\partial\left(\sum_i n_i\tau_i\right)=\sum n_i\sigma_i-\sum n_i \sigma=0\]from which we obtain the desired result.
The proof became somewhat lengthy for the sake of rigor, but the essential idea is that in a path-connected space \(X\), any two points can be connected by a path, and viewing this path as a \(1\)-simplex, these two points become the boundary of the \(1\)-simplex, so for \(B_0(X)=\im\partial_1\) we can regard these two points as equal.
Conversely, there are cases where we can compute the homology for all \(n\), namely when \(X\) is a point. In this case, regardless of the value of \(k\), a singular \(k\)-simplex \(\sigma_k:\Delta^k \rightarrow X\) is uniquely determined (that is, as a constant function), and considering formula (1), \(\partial_k\) is \(0\) when \(k\) is odd and sends \(\sigma_k\) to \(\sigma_{k-1}\) when \(k\) is even. That is, the following chain complex
\[\cdots\rightarrow \mathbb{Z} \overset{0}{\longrightarrow}\mathbb{Z}\overset{\approx}{\longrightarrow} \mathbb{Z}\overset{0}{\longrightarrow}\mathbb{Z}\rightarrow0\]is the chain complex of singular simplices, and thus we obtain the following.
Proposition 11 For a one-point space \(X\), \(H_0(X)\cong \mathbb{Z}\) and \(H_k(X)\cong 0\) for all \(k>0\).
However, of course, the most important property one might say 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 in 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 function \(f:X\rightarrow Y\), there exists a chain map \(C_\bullet(f):C_\bullet(X)\rightarrow C_\bullet(Y)\). Naturally, we can define \(C_\bullet(f)\) by the formula
\[C_\bullet(f):\sigma\mapsto f\circ\sigma\]and the key is to show that this is a chain map. That this is a chain map 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 define the following.
Definition 13 If \(H_n(X)\cong H_n(Y)\) holds for all \(n\), we say that two spaces \(X,Y\) are homologous.
References
[Hat] A. Hatcher, Algebraic Topology. Cambridge University Press, 2022.
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