Let \(X\) be the space obtained from the 2-simplex \(\Delta^2\) by identifying all three vertices together. Describe a \(\Delta\)-complex structure on the space \(X\) and compute its homology groups.
Solution
Write the vertices of the original simplex as \(v_0,v_1,v_2\), and let \(v\) denote their common image in \(X\). Orient the three original edges by \[ a=[v_0,v_1], \qquad b=[v_1,v_2], \qquad c=[v_0,v_2]. \]
1. The quotient \(X\) has a \(\Delta\)-complex structure with one \(0\)-simplex \(v\), three \(1\)-simplices \(a,b,c\), and one \(2\)-simplex \(\sigma\).
The quotient identifies only the three vertices of the standard simplex. The interior of each original edge is unaffected and maps injectively to \(X\), so the three edge interiors give three distinct open \(1\)-simplices whose two endpoints are all attached to the single vertex \(v\). The interior of \(\Delta^2\) is likewise unaffected and gives one open \(2\)-simplex \(\sigma\).
The quotient map restricted to the closed standard \(2\)-simplex is a characteristic map for \(\sigma\); each face is one of the three stated \(1\)-simplices, and their vertices all map to \(v\). These identifications are allowed in a \(\Delta\)-complex, so the stated cells form the required \(\Delta\)-complex structure.
2. Its simplicial chain groups are\[ C_2(X)=\mathbb Z\langle\sigma\rangle, \qquad C_1(X)=\mathbb Z\langle a,b,c\rangle, \qquad C_0(X)=\mathbb Z\langle v\rangle, \]with\[ \partial_1=0 \qquad\text{and}\qquad \partial_2(\sigma)=a+b-c. \]
Every \(1\)-simplex begins and ends at the same vertex \(v\), so \[ \partial_1(a)=\partial_1(b)=\partial_1(c)=v-v=0. \]
Orient \(\sigma\) as \([v_0,v_1,v_2]\). The simplicial boundary formula gives \[ \begin{aligned} \partial_2[v_0,v_1,v_2] &=[v_1,v_2]-[v_0,v_2]+[v_0,v_1]\\ &=b-c+a\\ &=a+b-c. \end{aligned} \]
3. We have\[ H_2(X)=0. \]
The homomorphism \[ \partial_2:\mathbb Z\longrightarrow\mathbb Z^3 \] sends \(1\) to the nonzero vector \[ (1,1,-1). \] Hence \(\partial_2\) is injective, so \[ H_2(X)=\ker\partial_2=0. \] There are no \(3\)-simplices.
4. We have\[ H_1(X)\cong\mathbb Z^2. \]
By 2, \[ \ker\partial_1=C_1(X)=\mathbb Z^3 \] and \[ \operatorname{im}\partial_2 = \mathbb Z\langle a+b-c\rangle. \] Thus \[ H_1(X) \cong \mathbb Z^3/\mathbb Z(1,1,-1). \] The vector \((1,1,-1)\) is primitive because its coordinates have greatest common divisor \(1\). Explicitly, the elements \[ a, \qquad b, \qquad d=a+b-c \] form a basis of \(\mathbb Z^3\): indeed \(c=a+b-d\), so they generate, and the change-of-basis matrix has determinant \(-1\). Quotienting by \(\mathbb Z d\) therefore leaves the free group generated by the classes of \(a\) and \(b\). Hence \[ H_1(X)\cong\mathbb Z^2. \]
5. We have\[ H_0(X)\cong\mathbb Z, \qquad H_n(X)=0\quad(n\ge3). \]
The \(1\)-skeleton has one vertex, so \(X\) is connected; equivalently, since \(\partial_1=0\) and \(C_0(X)=\mathbb Z\), the chain computation gives \[ H_0(X)=C_0(X)/\operatorname{im}\partial_1\cong\mathbb Z. \] There are no simplices in dimensions at least \(3\), so all higher chain groups and hence all homology groups \(H_n\) for \(n\ge3\) vanish.
6. Therefore\[ H_n(X)\cong \begin{cases} \mathbb Z, & n=0,\\ \mathbb Z^2, & n=1,\\ 0, & n\ge2. \end{cases} \]
Combine <1>3–<1>5.