Showing $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}/(1,-1,0)\mathbb{Z}+(0,1,1)\mathbb{Z}+(1,0,-1)\mathbb{Z}\cong \mathbb{Z}$

Question

I have to prove that if $V_K = \{v_0, v_1, v_2\}$ and $K = \{\{v_0\}, \{v_1\}, \{v_2\}, \{v_0, v_1\}, \{v_0, v_2\}, \{v_1, v_2\}\}$ then $H_q(K, \mathbb{Z})\cong \mathbb{Z}$ for $q = 0, 1$.

Already I proved that for $q=1$. For $H_0(K, \mathbb{Z})=\ker\delta_0/\operatorname{Im}\delta_1$, $\delta_0 : C_0(K)\to 0$, so it is obvious that $\ker\delta_0 = C_0(K)$. Also,

$$\operatorname{Im}\delta_1 = \{(a + c)v_0 + (-a + b)v_1 + (b - c)v_2 \mid a, b, c \in \mathbb{Z}\}$$

according to the definition of $\delta_1$.

Finally, this set can be written as follows:

$$\operatorname{Im}\delta_1 = (1, -1, 0)\mathbb{Z} + (0, 1, 1)\mathbb{Z} + (1, 0, -1)\mathbb{Z}$$

so the only thing I have to prove is that

$$\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}/(1,-1,0)\mathbb{Z}+(0,1,1)\mathbb{Z}+(1,0,-1)\mathbb{Z}\cong \mathbb{Z}.$$

All I can say is that the vectors $e_i$ don't belongs to this subset.

Answer 1

I believe you got a sign wrong in you expression for $\operatorname{Im}\delta_1$. You should have

$$\operatorname{Im}\delta_1 = \{(a + c)v_0 + (-a + b)v_1 + (\color{red}{-b} - c)v_2 \mid a, b, c \in \mathbb{Z}\}$$

as $\delta_1\{v_0, v_2\} = 1\{v_0\} \color{red}{- 1}\{v_2\}$, so $\operatorname{Im}\delta = (1, -1, 0)\mathbb{Z} + (1, 0, -1)\mathbb{Z} + (0, 1, -1)\mathbb{Z}$.

Now note that the group homomorphism $\varphi : \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z} \to \mathbb{Z}$ given by $\varphi(x, y, z) = x + y + z$ is surjective and $\operatorname{Im}\delta_1 \subseteq \ker\varphi$. If $(x, y, z) \in \ker\varphi$, then $x + y + z = 0$ so $z = -x - y$. As

$$(x, y, z) = (x, y, -x - y) = x(1, 0, -1) + y(0, 1, -1) \in \operatorname{Im}\delta_1,$$

we see that $\ker\varphi \subseteq\operatorname{Im}\delta_1$, and therefore $\ker\varphi = \operatorname{Im}\delta_1$.

So by the first isomorphism theorem,

$$\frac{\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}}{(1, -1, 0)\mathbb{Z} + (1, 0, -1)\mathbb{Z} + (0, 1, -1)\mathbb{Z}} \cong \mathbb{Z}.$$

Answer 2

The quotient is not equal to $\Bbb Z$: it is equal to $\Bbb Z/2$. In particular, to correct the comment by p Groups:

$f(1, -1, 0)=0$ implies $a-b=0$, so $a=b$. Similarly, $a-c=0$ implies $a=c$. Finally, $b+c=0$ implies $2a=0$. So the quotient is just $a\Bbb Z/2a\Bbb Z$, for example with representatives $\{(0, 0, 0), (1, 0, 0)\}$.

In particular, $(0, 0, 1) = (1, 0, 0) - (1, 0, -1)$, and $(0, 1, 0) = (1, 0, 0) - (1, -1, 0)$.

Answer 3

Performing elementary row and column operations you can bring your matrix of vectors to the form $$\begin{pmatrix}2&0&0\\0&1&0\\0&0&1\end{pmatrix}$$ so the quotient is $\Bbb Z_2$. In the corrected answer by Michael Albanese, you would end up with a $0$ instead of a $2$ (carry out the calculation yourself), so the quotient is indeed $\Bbb Z$. Note that the first homology group is zero, not $\Bbb Z$, unless the 2-simplex is not supposed to be there.