Simpler way of computing the first homology group of $\Delta^4$ (without the interior)?

Question

Here's my attempt. The first part (computing $Z_1$) feels like it's far too complicated. First of all - is my work correct? And second, is there a simpler way of doing this?

Let's label the $0$-simplices $(v_1,...,v_5)$, $1$-simplices $(e_1,...,e_{10})$. Now, we are looking for the following chain complex: $$C_2\overset{\partial_2}\to C_1\overset{\partial_1}\to C_0.$$ Here, $$C_0=\mathbb{Z} \langle v_1,v_2,v_3,v_4,v_5 \rangle $$ $$C_1=\mathbb{Z} \langle v_1v_2 \rangle \oplus\mathbb{Z} \langle v_1v_3 \rangle \oplus\mathbb{Z} \langle v_1v_4 \rangle \oplus\mathbb{Z} \langle v_1v_5 \rangle \oplus\mathbb{Z} \langle v_2v_3 \rangle \\ \oplus\mathbb{Z} \langle v_2v_4 \rangle \oplus\mathbb{Z} \langle v_2v_5 \rangle \oplus\mathbb{Z} \langle v_3v_4 \rangle \oplus\mathbb{Z} \langle v_3v_5 \rangle \oplus\mathbb{Z} \langle v_4v_5 \rangle $$ $$C_2=\mathbb{Z} \langle v_1v_2v_3 \rangle \oplus\mathbb{Z} \langle v_1v_2v_4 \rangle \oplus\mathbb{Z} \langle v_1v_2v_5 \rangle \oplus\mathbb{Z} \langle v_1v_3v_4 \rangle \oplus\mathbb{Z} \langle v_1v_3v_5 \rangle \\ \oplus\mathbb{Z} \langle v_1v_4v_5 \rangle \oplus\mathbb{Z} \langle v_2v_3v_4 \rangle \oplus\mathbb{Z} \langle v_2v_3v_5 \rangle \oplus\mathbb{Z} \langle v_2v_4v_5 \rangle \oplus\mathbb{Z} \langle v_3v_4v_5 \rangle $$ $\partial_1$ does the following maps: $$e_1 \mapsto v_2-v_1$$ $$e_2 \mapsto v_3-v_1$$ $$e_3 \mapsto v_4-v_1$$ $$e_4 \mapsto v_5-v_1$$ $$e_5 \mapsto v_3-v_2$$ $$e_6 \mapsto v_4-v_2$$ $$e_7 \mapsto v_5-v_2$$ $$e_8 \mapsto v_4-v_3$$ $$e_9 \mapsto v_5-v_3$$ $$e_{10} \mapsto v_5-v_4$$ Now, to compute $H_1$ we need to find $Z_1=\ker(\partial_1)$. Therefore, we need to find the linear combinations of $\partial_1(e_1),...,\partial_1(e_{10})$ such that $$\partial_1\Big( w_1(e_1)+....+w_{10}(e_{10})\Big) =0$$ $$w_1(v_2-v_1)+...+w_{10}(v_5-v_4)=0.$$ We can re-write this as $$v_1(-w_1-w_2-w_3-w_4)+...+v_5(w_4+w_7+w_9+w_{10})=0$$ which is equivalent to setting the following matrix to zero (columns represent the $w_i$s, rows represent the $v_i$s):

$$\begin{pmatrix} -1 & -1 & -1 & -1 & 0 & 0 &0 &0 & 0&0 \\ 1& 0 &0 &0 & -1 &-1 &-1 &0 &0 &0 \\ 0& 1 & 0 &0 &1 &0 &0 &-1 &-1 &0 \\ 0& 0&1 &0 &0 &1 & 0 &1 &0 &-1 \\ 0&0 & 0&1 & 0 & 0 & 1 & 0 &1 &1 \end{pmatrix}$$ Putting it into row echelon form: $$\begin{pmatrix} 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0& 1 &1 &1 & 1 &1 &1 &0 &0 &0 \\ 0& 0 & 1 &1 &0 &1 &1 &1 &1 &0 \\ 0& 0& 0 &1 &0 &0 & 1 &0 &1 &1 \\ 0&0 & 0&0 & 0 & 0 & 0 & 0 &0 &0 \end{pmatrix}$$ $$\begin{pmatrix} 1 & 0 & 0 & 0 & -1 & -1 & -1 & 0 & 0 & 0 \\ 0& 1 & 0 &0 & 1 &0 &0 &-1 &-1 &0 \\ 0& 0 & 1 &0 &0 &1 &0 &1 &0 &-1 \\ 0& 0& 0 &1 &0 &0 & 1 &0 &1 &1 \\ 0&0 & 0&0 & 0 & 0 & 0 & 0 &0 &0 \end{pmatrix}$$ Let's call the $5^{th}$th column $r_1$, the $6^{th}$ one $r_2$, the $7^{th}$ one $r_3$, the $8^{th}$ one $r_4$, the $9^{th}$ one $r_5$, the $10^{th}$ one $r_6$. So now $$r_1\begin{pmatrix} 1 \\ -1 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}+ r_2\begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}+ r_3\begin{pmatrix} 1 \\ 0\\ 0 \\ -1 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}+ r_4\begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}+ r_5\begin{pmatrix} 0 \\ 1 \\ 0 \\ -1\\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}+ r_6\begin{pmatrix} 0 \\ 0 \\ 1 \\ -1\\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}= \begin{pmatrix} w_1 \\ w_2 \\ w_3 \\ w_4\\ w_5 \\ w_6 \\ w_7 \\ w_8 \\ w_9 \\ w_{10} \end{pmatrix}$$ Solutions: multiples of cycles: $$w_1-w_2+w_5$$ $$w_1-w_3+w_6$$ $$w_1-w_4+w_7$$ $$w_2-w_3+w_8$$ $$w_2-w_4+w_9$$ $$w_3-w_4+w_{10},$$ where we associate each $w_i$ with the corresponding edge $e_i$. So $Z_1=\mathbb{Z}^6$. Now we need to find $B_1$, which is the image of $\partial_2$. The boundaries of the $2$-simplices can all be written as multiples of the cycles listed above. Hence, $B_1=Z_1$ and their quotient must be the trivial group $0$.

Answer 1

From your title, I am going to assume that your intention is to compute the homology of $\Delta^4 - \text{interior}(\Delta^4)$, although you did not mention that again in the text of the question.

One thing about homology is, if you find yourself doing torturous computations, especially with such a fundamental example, probably there is an easier way.

$\Delta^4$, being contractible, has trivial homology in all dimensions except in dimension $0$ where its homology is $\mathbb{Z}$ since it is connected.

$\Delta^4 - \text{interior}(\Delta^4)$ is equal to the 3-skeleton of $\Delta^4$. So, the chain complexes for $\Delta^4$ and for $\Delta^4 - \text{interior}(\Delta^4)$ are identical in dimensions $0$ through $3$, from which it follows that their homologies are isomorphic in dimensions $0$ $1$, and $2$. Thus, the homologies of $\Delta^4 - \text{interior}(\Delta^4)$ are $\mathbb{Z}$ in dimension $0$ and trivial in dimensions $1$ and $2$.

Also, the homologies of $\Delta^4 - \text{interior}(\Delta^4)$ are all zero from dimension $4$ and up, it being 3-dimensional.

That leaves dimension $3$. Each 3-simplex in $\Delta^4$ has ordered vertex set obtained from the list $(v_0,v_1,v_2,v_3,v_4)$ by removing one vertex $v_i$; give it a coefficient $+1$ if $i$ is even, $-1$ if $i$ is odd. This defines a $3$-cycle, as one can check. Furthermore, every $3$-cycle is a scalar multiple of this one, because every $2$-simplex is on the boundary of exactly two $3$-simplices, and so changing the coefficient on one 3-simplex by adding the integer $k$ forces the exact same change on each adjacent $3$-simplex by also adding $k$. Finally, since there are no 4-simplices in $\Delta^4 - \text{interior}(\Delta^4)$ its group of 3-cycles is isomorphic to its 3-dimensional homology. Thus, $$H_3(\Delta^4 - \text{interior}(\Delta^4)) = \mathbb{Z} $$

Answer 2

The following answer is not the calculation of the homology of $\Delta^4\setminus int ( \Delta^4)$, but of $\Delta^4$. The correct answer can be found under the title "edit".

You can find in Hatcher (p. 111) something like the following statement:

Corollary: If $X$ and $Y$, two topological spaces, are homotopic equivalent, then $H_n(X)\cong H_n(Y)$.

Now you have to know this

Lemma: Any convex subset $X\subseteq \mathbb{R}^n$ satisfies $H_n(X,\mathbb{Z})=\begin{cases} \mathbb{Z}\mbox{ for } n=0,\\ 0 \mbox{ else.}\end{cases}$

Proof: Choose a point $p\in X$ and set $G\colon X\times I\to X,$ $(x,t)\mapsto tp+(1-t)x$.

$G$ is an homotopy from $G|_{X\times\{0\}}=id_X$ to $G|_{X\times\{1\}}=const_{p}$ and $G|_{\{p\}\times I}=id$, hence it is a deformation retraction. $\square$

Remark:

1.) The $const_p$ maps any element to $p$.

2.) Obviously $H_n(pt.,\mathbb{Z})=\begin{cases} \mathbb{Z}\mbox{ for } n=0,\\ 0 \mbox{ else.}\end{cases}$

3.) If $A\subseteq X$ is a deformation retract, then $A\simeq X$ are homotopic equivalent.

Another strategy:

Any standard $n$-simplex is homeomorphic to the $n$-ball. Then calculate the homology group of the $n$-ball.

Edit

Sorry for this late correction! The OP wanted to calculate $\Delta^4\setminus int (\Delta^4) $. Here's an approach. I will show a more general case:

The standard $ q $-simplex is defined as $$\Delta^q=\{(t_0,..., t_q)|\sum_{i=0}^qt_i=1, t_i\in I\} $$ Now $$\{(t_0,..., t_q)|\sum_{i=0}^qt_i=1, t_i\in I\mbox { and at least one }t_i=0 \}=\partial\Delta^q=\bar {\Delta^q}\setminus int (\Delta^q)=\Delta^q\setminus int (\Delta^q),$$ because $\Delta^q $ is closed.

$\partial \Delta^q $ is homeomorphic to $ \mathbb{S}^{q-1} $ and homeomorphic spaces have isomorphic homology groups.

The problem now is just the calculation of the homology of the sphere $\mathbb {S}^n $.

The calculation can be done for example with Mayer-Vietoris:

Choose $ U=\mathbb {S}^n\setminus N $ and $ V=\mathbb {S}^n\setminus S $, where $ N$ and $ S $ are the north and south pole. $ U $ and $ V $ are both homeomorphic to $\mathbb {R}^n $ by stereographic projection, which has obviously the homology group of a point (deformation retraction). The intersection $ U\cap V $ is homotopy equivalent to $ \mathbb {S}^{n-1} $ (deformation retraction).

We have the following exact sequence for $ p> 0$:

$$\underbrace {H_{p+1}(U)\oplus H_{p+1}(V)}_{=0}\longrightarrow H_{p+1}(\mathbb {S}^n)\overset {\sim}{\longrightarrow} \underbrace {H_{p}(U\cap V)}_{=H_p (\mathbb {S}^{n-1})}\longrightarrow \underbrace {H_p (U)\oplus H_p (V)}_{=0} $$ Because of exactness the middle map is a homomorphism between $ H_{p+1}(\mathbb{S}^n)\cong H_p (\mathbb {S}^{n-1}) $. Since $ H_0$ "counts" path-connected components, $ H_0 (\mathbb {S}^0)=\mathbb {Z}^2$ and $ H_0 (\mathbb {S}^n)=\mathbb {Z} $ for $ n> 0$.

Now for $ n=1$ the map $\mathbb {Z}\oplus \mathbb {Z}=H_0 (\mathbb {S}^0)\longrightarrow H_0 (\mathbb {R})\oplus H_0 (\mathbb {R})=\mathbb {Z}\oplus \mathbb {Z}$ is given by $(x, y)\mapsto (x+y, x+y) $.

By exactness we deduce $ H_p (\mathbb {S}^1)=\begin {cases}\mathbb {Z}, p=0,1\\0, \mbox { else}\end {cases} $.

By the homeomorphism before we can deduce inductively $ H_p (\mathbb {S}^n)=\begin {cases}\mathbb {Z}, p=0,n\\0, \mbox { else}\end {cases} $ for $ n\in \mathbb {N} $.

All in all $$ H_p (\Delta^q\setminus int (\Delta^q))=H_p (\mathbb {S}^{q-1})=\begin {cases} \mathbb {Z}, p=0, q-1\\0, \mbox { else}\end {cases}. $$