Check my working: how could $H_1^\Delta(T)=\mathbb{Z}_2$?

Question

My attempt for the following question:

My attempt:

$H_1^\Delta(T)=\frac{\ker\partial_1}{\text{im }\partial_2}=\frac{<c,b-a>}{<a-b-c,a-b+c>}=\frac{\mathbb{Z}\oplus\mathbb{Z}}{\mathbb{Z}\oplus\mathbb{Z}}=0$, since the space generated by two generators is isomorphic to $\mathbb{Z}\oplus\mathbb{Z}$.

Is my working correct?

The question is taken from Hatcher's page 106-107.

My doubts:

1. I understand almost every of the given solution, but what confuses me is how can the fact the $\text{im }\partial_2$ is an index-two subgroup of $\ker\partial_1$ implies $H_1^\Delta(T)=\mathbb{Z}_2$?

2. It is written that $\partial_2$ is injective, my other question is whether the boundary homomorphism $\partial_i$ is always injective?

Could someone please give some light on this? Helps are really appreciated. Thanks!

Answer 1

If you play with the algebra, $\langle c, b-a \rangle = \langle c, b-a-c \rangle$ and $\langle a-b-c,a-b+c \rangle = \langle a-b-c,b-a-c \rangle = \langle -2c,b-a-c\rangle = \langle 2c,b-a-c \rangle$ so the result is: $$H_1(\mathbb{R}P^2) = \frac{ \langle c, b-a-c \rangle}{\langle 2c,b-a-c \rangle} \cong \mathbb{Z}_2$$

$\partial_i$ is not always injective. If so, ker $\partial_1$ would be zero in this case :)

Answer 2

Your mistake is that, although it is correct that the submodule you're taking the quotient by is isomorphic to $\mathbb Z\oplus\mathbb Z$, you don't know how it sits inside the bigger $\mathbb Z\oplus\mathbb Z$. Take as an example $3\mathbb Z$ inside $\mathbb Z$, which is isomorphic to $\mathbb Z$ but certainly $\mathbb Z/3\mathbb Z$ is not zero. As explained in the other answer, the submodule you're looking at is the direct sum $2\mathbb Z\oplus\mathbb Z$ inside $\mathbb Z\oplus\mathbb Z$, and now you can safely assert that the quotient is $\mathbb Z/2\mathbb Z$.

The canonical algorithm that elucidates the isomorphism class of a submodule $M$ of $\mathbb Z^n$ along with the isomorphism class of the quotient $\mathbb Z^n/M$ in terms of a set of generators is the Smith normal form. In particular, this should help you compute the cellular homology of spaces in low dimensions, where you can explicitly give the generators of kernels and images.

Answer 3

Your chain complex is

The equations \begin{align*} \partial_2(U) &= -a+b+c & \partial_1(a) &= -v+w\\ \partial_2(L) &= a-b+c & \partial_1(b) &= -v+w \\ &&\partial_1(c) &= 0 \end{align*}

allow us to view the boundary maps as matrices \begin{align*} \partial_2 &= \left[\begin{array}{rr} -1 & 1 \\ 1 & -1 \\ 1 & 1 \end{array}\right] & \partial_1 &= \left[\begin{array}{rrr} -1 & -1 & 0 \\ 1 & 1 & 0 \end{array}\right] \end{align*}

We can now use Smith normal form to quickly infer the homology. Note that the Smith decompositions of $\partial_1$ and $\partial_2$ are $D_k=L_k\partial_k R_k$ where \begin{align*} D_2 &= \left[\begin{array}{rr} 1 & 0 \\ 0 & 2 \\ 0 & 0 \end{array}\right] & L_2 &= \left[\begin{array}{rrr} 0 & 0 & 1 \\ -1 & 0 & -1 \\ 1 & 1 & 0 \end{array}\right] & R_2 &= \left[\begin{array}{rr} 1 & 1 \\ 0 & -1 \end{array}\right] \\ D_1 &= \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] & L_1 &= \left[\begin{array}{rr} -1 & 0 \\ 1 & 1 \end{array}\right] & R_1 &= \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right] \end{align*}

The equation $\partial_2R_2=L_2^{-1}D_2=$ is equivalent to \begin{align*} \partial_2(U) &= -a+b+c & \partial_2(U-L)&=2\,(-a+b) \end{align*} Thus $\DeclareMathOperator{im}{im}\im(\partial_2)=\Bbb Z(-a+b+c)\oplus\Bbb Z(2\,(-a+b))$.

Next, note that $\partial_1$ has a rank-two kernel and that $$ \partial_1(-a+b+c)=\partial_1(-a+b)=9 $$ Thus $\ker(\partial_1)=\Bbb Z(-a+b+c)\oplus\Bbb Z(-a+b)$.

Hence our homology is $$ H_1(X)=\frac{\Bbb Z(-a+b+c)\oplus\Bbb Z(-a+b)}{\Bbb Z(-a+b+c)\oplus\Bbb Z(2\,(-a+b))} \simeq \Bbb Z/2\Bbb Z $$