Exercises:Saul - Algebraic Topology - 7/Exercise 7.6

Exercises

Exercises 7.6

Question

1. Compute the singular homology groups of $T^2:\eq\mathbb{S}^1\times\mathbb{S}^1$ and of $X:\eq\mathbb{S}^1\vee\mathbb{S}^1\vee\mathbb{S}^2$
2. Prove that $T^2$ and $X$ are not homotopy equivalent spaces

Solutions

Part I

$\Delta$-complex for $X$
$\xymatrix{ \bullet_{v_0} \ar[dd]_\alpha \ar[dr]^\beta & & \bullet_{v_3} \ar@2{->}[dr]^b \ar@2{->}[dl]_a & & \bullet_{v_5} \ar[dd]^z \\ & \bullet_{v_2} \ar@2{<-}[dr]_a \ar@{}[rr]|(.3){\LARGE{\mathbf{A} } }|(.7){\LARGE{\mathbf{ B} } } & & \bullet_{v_4} \ar@{<-}[ur]^x \\ \bullet_{v_1} \ar[ur]_\gamma & & \bullet_{v_3} \ar@2{->}[ur]_b \ar@2{->}[uu]^c_c & & \bullet_{v_6} \ar[ul]^y }$
$0$-simplices:	$v_0,v_1,v_2,v_3,v_4,v_5,v_6$	(7)
$1$-simplices:	$\alpha,\beta,\gamma,a,b,c,x,y,z$	(9)
$2$-simplices:	$A,B$	(2)

The homology groups of the $2$-torus have been computed in previous assignments, the result was:

• $H_0^\Delta(T^2)\cong\mathbb{Z}$
• $H_1^\Delta(T^2)\cong\mathbb{Z}\oplus\mathbb{Z}\eq:\mathbb{Z}^2$
• $H_2^\Delta(T^2)\cong\mathbb{Z}$
• $H_n^\Delta(T^2)\cong 0$ for $n\in\{3,\ldots\}\subset\mathbb{N}$

To calculate the homology groups of $X$ we must start by computing the boundary of the generators (which are extended to group homomorphisms on free Abelian group by the characteristic property of free Abelian groups):

• $2$-simplices:

  $\partial_2(A)\eq c$, $\partial_2(B)\eq c$

• $1$-simplices:

  $\partial_1(\alpha)\eq v_1-v_0$, $\partial_1(\beta)\eq v_2-v_0$, $\partial_1(\gamma)\eq v_2-v_1$

  $\partial_1(a)\eq v_2-v_3$, $\partial_1(b)\eq v_4-v_3$, $\partial_1(c)\eq 0$

  $\partial_1(x)\eq v_4-v_5$, $\partial_1(y)\eq v_4-v_6$, $\partial_1(z)\eq v_6-v_5$

• Note that $\partial_0:(\text{anything})\mapsto 0$, so we do not mention in here, it is also obvious that the entire of the domain is the kernel of $\partial_0$ from this definition.

Now the images and kernels:

• $\text{Im}(\partial_2)\eq \langle c\rangle$ - by inspection
• $\text{Ker}(\partial_2)\eq \langle A-B \rangle$ - by inspection
• $\text{Im}(\partial_1)\eq \langle v_0-v_1,\ v_0-v_2,\ v_2-v_3,\ v_3-v_4,\ v_4-v_5,\ v_6-v_4\rangle$
  • Calculated by starting with just the first vector, $\partial_1(\alpha)$, going over each of the remaining vectors and adding it to this linearly-independent set if said vector is not a linear combination of what we have in this set.
  • $\partial_1(\gamma)\eq\partial_1(\beta)-\partial_1(\beta)$, then we see $\partial_1(c)$ is the identity element, $0$, so cannot be in a basis set for obvious reasons, and $\partial_1(z)\eq\partial_1(x)-\partial_1(y)$, hence the chosen basis consists of all but these.
• $\text{Ker}(\partial_1)\eq\langle\alpha-\beta+\gamma, c, x-y-z\rangle$
  • Computed by "RReffing in $\mathbb{Z}$", I may upload a picture of these matrices, but as it is $10$ columns by $7$ rows I am not eager to type both the starting matrix and it's reduced form out.
• $\text{Ker}(\partial_0)\eq\langle v_0,v_1,v_2,v_3,v_4,v_5,v_6\rangle$, as discussed above and from the definition of $\partial_0$

Note that for any higher values:

• $\text{Ker}(\partial_n)\eq 0$ for $n\ge 3$, $n\in\mathbb{N}$, as these groups are trivial groups, and a group homomorphism must map the identity to the identity, couple this with the domain of $\partial_n$ has one element and the result follows.
• $\text{Im}(\partial_n)\eq 0$ for $n\ge 3$, $n\in\mathbb{N}$, as the image of a trivial group must be the identity element of the co-domain group.

Homology groups of $X$

• $$H_2^\Delta(X):\eq\frac{\text{Ker}(\partial_2)}{\text{Im}(\partial_3)}\eq\frac{\langle A-B \rangle}{0}\cong \langle A-B \rangle\cong\mathbb{Z}$$
• $$H_1^\Delta(X):\eq\frac{\text{Ker}(\partial_1)}{\text{Im}(\partial_2)}\eq\frac{\langle\alpha-\beta+\gamma, c, x-y-z\rangle}{\langle c\rangle}\cong\langle\alpha-\beta+\gamma,x-y-z\rangle\cong\mathbb{Z}^2$$
• $$H_0^\Delta(X):\eq\frac{\text{Ker}(\partial_0)}{\text{Im}(\partial_1)}\eq\frac{\langle v_0,v_1,v_2,v_3,v_4,v_5,v_6\rangle}{\langle v_0-v_1,\ v_0-v_2,\ v_2-v_3,\ v_3-v_4,\ v_4-v_5,\ v_6-v_4\rangle} \cong\mathbb{Z}$$
  • We compute this (if we want to go the hard way) by re-writing $\langle v_0,v_1,v_2,v_3,v_4,v_5,v_6\rangle$ step by step until it looks like $\langle v_0-v_1,\ v_0-v_2,\ v_2-v_3,\ v_3-v_4,\ v_4-v_5,\ v_6-v_4\rangle$, for example:
    • $\text{Span}(v_0,v_1)\eq\text{Span}(v_0-v_1,v_1)$, we do this by noticing that each "vector" (linear combination) in the span can be made from members in the other span, and visa-versa, so these spans are equal (we show $\subseteq$ and $$\supseteq$$ ), and just keep going until the numerator looks like the denominator with some extra terms
• $$H_n^\Delta(X)\eq\frac{\text{Ker}(\partial_n)}{\text{Im}(\partial_{n+1})}\eq\frac{0}{0}\cong 0$$ for $n\ge 3$ and $n\in\mathbb{N}$ as discussed above

Singular homology groups

By Hatcher Theorem 2.27 found on page 128.6 (with $A\eq\emptyset$) we see that the singular homology groups, $H_n(S)$ for a topological space $S$ are isomorphic to the simplicial (or delta-complex specifically) homology groups. That is:

• $\forall n\in\mathbb{N}_0[H_n^\Delta(S)\cong H_n(S)]$, so we see:
  • $H_0(X)\cong H_0^\Delta(X)\cong\mathbb{Z}$ and $H_0(T^2)\cong H_0^\Delta(T^2)\cong\mathbb{Z}$, then
  • $H_1(X)\cong H_1^\Delta(X)\cong\mathbb{Z}\oplus\mathbb{Z}\cong\mathbb{Z}^2$ and $H_1(T^2)\cong H_1^\Delta(T^2)\cong\mathbb{Z}\oplus\mathbb{Z}\cong\mathbb{Z}^2$, then
  • $H_2(X)\cong H_2^\Delta(X)\cong\mathbb{Z}$ and $H_2(T^2)\cong H_2^\Delta(T^2)\cong\mathbb{Z}$, and lastly
  • for $n\in\mathbb{N}$ and $n\ge 3$:
    • $H_n(X)\cong H_n^\Delta(X)\cong 0$ and $H_n(T^2)\cong H_n^\Delta(T^2)\cong 0$

Part II

Observe that both $X$ and $T^2$ are path-connected topological spaces. As a result we will write $\pi_1(X)$ or $\pi_1(T^2)$ for their fundamental groups (respectively), knowing that for a path-connected space the fundamental groups based anywhere are isomorphic to each other.

Note that:

• $\pi_1(T^2)\cong \mathbb{Z}^2$ and
• $\pi_1(X)\cong \mathbb{Z}\ast\mathbb{Z}$

Recall the following:

Claim:
Hatcher - p37	Proposition 1.18: if $\varphi:X\rightarrow Y$ is a homotopy equivalence, then the induced homomorphism $\varphi_*:\pi_1(X,x_0)\rightarrow\pi_1(Y,\varphi(x_0))$ is an isomorphism for all $x_0\in X$

Specifically notice that: $\text{Homotopy equivalent}\implies\text{isomorphic fundamental groups}$, we invoke the contrapositive, which is logically equivalent (if and only if) to this:

• $\neg(\text{isomorphic fundamental groups})\implies\neg(\text{Homotopy equivalent})$

or

• fundamental groups are not isomorphic $\implies$ the spaces are not homotopy equivalent

Clearly $\mathbb{Z}^2\ncong\mathbb{Z}\ast\mathbb{Z}$, thus $X \not\simeq T^2$, as required (where $X\simeq T^2$ would denote that $X$ (the space) is homotopy equivalent to $T^2$ and the line through it means "not", like $\eq$ and $\neq$)

Appendix

TODO: Lee - Topological manifolds - page 255. Computing the fundamental group of a wedge product