I am self studying some algebraic topology and came over the following exercise:
Take an octagon whose edges are labelled α, β, α, α, α, β, γ, γ, as one goes around clockwise (and all edges are oriented clockwise, too). Let X be the space obtained by identifying edges which are labelled the same. Compute the groups $H(X)$ and $H(X;\mathbb{F}_2)$.
I explicitly constructed a cellular chain complex
$$0 \rightarrow \mathbb{Z} \overset{\partial_1}{\rightarrow} \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \overset{\partial_0}{\rightarrow} \mathbb{Z} \rightarrow 0$$
Where $\partial_0 = 0$ and $\partial_1(\sigma) = 4 \alpha + 2 \beta + 2 \gamma$ where $\sigma$ is the single 2 cell of $X$.
From this I obtained the following homology groups in integer coefficients.
$$
H_0(X) \cong \mathbb{Z} \\
$$
$$
H_1(X) \cong \mathbb{Z}/4\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}
$$
$$
H_2(X) \cong 0
$$
Further I applied UCT to compute the homology with mod 2 coefficients.
$$
H_0(X;\mathbb{F}_2) \cong H_0(X) \otimes \mathbb{F}_2 \cong \mathbb{F}_2
$$
$$
H_1(X;\mathbb{F}_2) \cong H_1(X) \otimes \mathbb{F}_2 \oplus Tor(H_0(X),\mathbb{F}_2) \cong \mathbb{F}_2 \oplus \mathbb{F}_2 \oplus \mathbb{F}_2
$$
$$
H_2(X;\mathbb{F}_2) \cong H_2(X) \otimes \mathbb{F}_2 \oplus Tor(H_1(X),\mathbb{F}_2) \cong \mathbb{F}_2 \oplus \mathbb{F}_2 \oplus \mathbb{F}_2
$$
However, trying to explicitly compute the mod 2 homology I get the following complex after tensoring with $\mathbb{F}_2$
$$0 \rightarrow \mathbb{F}_2 \overset{0}{\rightarrow} \mathbb{F}_2 \oplus \mathbb{F}_2 \oplus \mathbb{F}_2 \overset{0}{\rightarrow} \mathbb{F}_2 \rightarrow 0$$
This gives different homology groups in degree 2. Where have I gone wrong?
Best Answer
Your computation of $H_1(X)$ is incorrect. You took a quotient by $(4\mathbb{Z},2\mathbb{Z},2\mathbb{Z})$ instead of $(4,2,2)\mathbb{Z}$. The correct computation gives $H_1(X)\cong \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}\times \mathbb{Z}$ and then the computation by UCT gives the same answer.