# Exercises:Saul - Algebraic Topology - 7/Exercise 7.6

## Exercises

### Exercises 7.6

#### Question

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

#### Solutions

##### Part I
[ilmath]\Delta[/ilmath]-complex for [ilmath]X[/ilmath]
[ilmath]\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 }[/ilmath]
[ilmath]0[/ilmath]-simplices: [ilmath]v_0,v_1,v_2,v_3,v_4,v_5,v_6[/ilmath] (7)
[ilmath]1[/ilmath]-simplices: [ilmath]\alpha,\beta,\gamma,a,b,c,x,y,z[/ilmath] (9)
[ilmath]2[/ilmath]-simplices: [ilmath]A,B[/ilmath] (2)
The homology groups of the [ilmath]2[/ilmath]-torus have been computed in previous assignments, the result was:
• [ilmath]H_0^\Delta(T^2)\cong\mathbb{Z} [/ilmath]
• [ilmath]H_1^\Delta(T^2)\cong\mathbb{Z}\oplus\mathbb{Z}\eq:\mathbb{Z}^2[/ilmath]
• [ilmath]H_2^\Delta(T^2)\cong\mathbb{Z} [/ilmath]
• [ilmath]H_n^\Delta(T^2)\cong 0[/ilmath][Note 1] for [ilmath]n\in\{3,\ldots\}\subset\mathbb{N} [/ilmath]

Note: We have taken the wedge sum as [ilmath]\mathbb{S}^1\vee(\mathbb{S}^1\vee\mathbb{S}^2)[/ilmath] which is okay to do. In fact the wedge sum is even commutative (up to isomorphism, or homeomorphism as we call it in topology), we can also see this if we note that a wedge sum is an instance of a categorical coproduct. See the appendix at the bottom for details.

To calculate the homology groups of [ilmath]X[/ilmath] 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):

• [ilmath]2[/ilmath]-simplices:
[ilmath]\partial_2(A)\eq c[/ilmath], [ilmath]\partial_2(B)\eq c[/ilmath]
• [ilmath]1[/ilmath]-simplices:
[ilmath]\partial_1(\alpha)\eq v_1-v_0[/ilmath], [ilmath]\partial_1(\beta)\eq v_2-v_0[/ilmath], [ilmath]\partial_1(\gamma)\eq v_2-v_1[/ilmath]
[ilmath]\partial_1(a)\eq v_2-v_3[/ilmath], [ilmath]\partial_1(b)\eq v_4-v_3[/ilmath], [ilmath]\partial_1(c)\eq 0[/ilmath]
[ilmath]\partial_1(x)\eq v_4-v_5[/ilmath], [ilmath]\partial_1(y)\eq v_4-v_6[/ilmath], [ilmath]\partial_1(z)\eq v_6-v_5[/ilmath]
• Note that [ilmath]\partial_0:(\text{anything})\mapsto 0[/ilmath], so we do not mention in here, it is also obvious that the entire of the domain is the kernel of [ilmath]\partial_0[/ilmath] from this definition.

Now the images and kernels:

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

Note that for any higher values:

• [ilmath]\text{Ker}(\partial_n)\eq 0[/ilmath] for [ilmath]n\ge 3[/ilmath], [ilmath]n\in\mathbb{N} [/ilmath], as these groups are trivial groups, and a group homomorphism must map the identity to the identity, couple this with the domain of [ilmath]\partial_n[/ilmath] has one element and the result follows.
• [ilmath]\text{Im}(\partial_n)\eq 0[/ilmath] for [ilmath]n\ge 3[/ilmath], [ilmath]n\in\mathbb{N} [/ilmath], as the image of a trivial group must be the identity element of the co-domain group.
###### Homology groups of [ilmath]X[/ilmath]
• $H_2^\Delta(X):\eq\frac{\text{Ker}(\partial_2)}{\text{Im}(\partial_3)}\eq\frac{\langle A-B \rangle}{0}\cong \langle A-B \rangle\cong\mathbb{Z}$
• $H_1^\Delta(X):\eq\frac{\text{Ker}(\partial_1)}{\text{Im}(\partial_2)}\eq\frac{\langle\alpha-\beta+\gamma, c, x-y-z\rangle}{\langle c\rangle}\cong\langle\alpha-\beta+\gamma,x-y-z\rangle\cong\mathbb{Z}^2$
• $H_0^\Delta(X):\eq\frac{\text{Ker}(\partial_0)}{\text{Im}(\partial_1)}\eq\frac{\langle v_0,v_1,v_2,v_3,v_4,v_5,v_6\rangle}{\langle v_0-v_1,\ v_0-v_2,\ v_2-v_3,\ v_3-v_4,\ v_4-v_5,\ v_6-v_4\rangle} \cong\mathbb{Z}$
• We compute this (if we want to go the hard way) by re-writing [ilmath]\langle v_0,v_1,v_2,v_3,v_4,v_5,v_6\rangle[/ilmath] step by step until it looks like [ilmath]\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[/ilmath], for example:
• [ilmath]\text{Span}(v_0,v_1)\eq\text{Span}(v_0-v_1,v_1)[/ilmath], 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 [ilmath]\subseteq[/ilmath] 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 [ilmath]n\ge 3[/ilmath] and [ilmath]n\in\mathbb{N} [/ilmath] as discussed above

###### Singular homology groups

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

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

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

Note that:

• [ilmath]\pi_1(T^2)\cong \mathbb{Z}^2[/ilmath] and
• [ilmath]\pi_1(X)\cong \mathbb{Z}\ast\mathbb{Z} [/ilmath][Note 2]

Recall the following:

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

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

• [ilmath]\neg(\text{isomorphic fundamental groups})\implies\neg(\text{Homotopy equivalent})[/ilmath]

or

• fundamental groups are not isomorphic [ilmath]\implies[/ilmath] the spaces are not homotopy equivalent

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

#### Appendix

##### Computing the fundamental groups of wedge spaces

We assume the reader knows:

1. the Seifert-Van Kampen theorem, and
2. the definition of a non-degenerate base point of a topological space:
• [ilmath]p\in X[/ilmath] is a non-degenerate base point of [ilmath]X[/ilmath] if [ilmath]p[/ilmath] has an open neighbourhood that admits a strong deformation retraction onto [ilmath]p[/ilmath]
• Alec's note: it's not hard to see that actually this isn't that strict of a thing to ask for. Especially on some [ilmath]n[/ilmath]-sphere; which is, informally, "the same everywhere"

We then claim:

• Let [ilmath]\big((X_i,\mathcal{J}_i)\big)_{i\eq 1}^n[/ilmath] be a family of topological spaces and let [ilmath](p_i)_{i\eq 1}^n[/ilmath] be a family of non-degenerate base points of [ilmath]X_i[/ilmath] (for each [ilmath]i[/ilmath]), then the map:
• [ilmath]\Phi:\pi_1(X_1,p_1)\ast\cdots\ast\pi_1(X_n,p_n)\rightarrow\pi_1(X_1\vee\cdots\vee X_n,p_*)[/ilmath] is an isomorphism
• [ilmath]p_\ast[/ilmath] is the point in [ilmath]X_1\vee\cdots\vee X_n[/ilmath] that is the equivalence class which the [ilmath]p_i[/ilmath] are sent to (in this case the [ilmath]p_i[/ilmath] are all joined together by the wedge sum)
###### Fundamental group of [ilmath]\mathbb{S}^1\vee(\mathbb{S}^1\vee\mathbb{S}^2)[/ilmath]

Recall:

• [ilmath]\pi_1(\mathbb{S}^1)\cong\mathbb{Z} [/ilmath] and
• [ilmath]\pi_1(\mathbb{S}^2)\cong 1[/ilmath] (we now use [ilmath]1[/ilmath] to denote the trivial group)

It is easy to see that [ilmath]\mathbb{Z}\ast 1\cong\mathbb{Z} [/ilmath] by definition of the free product, as any word of the form [ilmath](n,1,m)[/ilmath] will be reduced to [ilmath](n,m)[/ilmath] then [ilmath]n+m[/ilmath])

So then [ilmath]\pi_1(\mathbb{S}^1\vee(\mathbb{S}^1\vee\mathbb{S}^2))[/ilmath] is just [ilmath]\mathbb{Z}\ast\mathbb{Z} [/ilmath] as required.

The pictures below show this pictorially (obviously, they're pictures)