Homology of $S^1 \times (S^1 \vee S^1)$

Question

I'm trying to solve question 2.2.9(b) in Hatcher's Algebraic Topology.

Question: Calculate the homology groups of $X=S^1 \times (S^1 \vee S^1)$.

My attempt: I try to use the Mayer-Vietoris sequence.

Let $A=S^1 \times (S^1 \vee \text{small bit}) =S^1\times S^1$ and $B=S^1 \times ( \text{small bit} \vee S^1)=S^1 \times S^1$. (Unsure how to express this, but hopefully this is clear enough.)

Then $A\cap B=S^1\times \{\text{point}\}=S^1$ and $A\cup B=X$.

The Mayer-Vietoris sequence then gives us an exact sequence in reduced homology as follows:

$$0 \to \mathbb{Z}^2 \to \tilde{H}_2(X) \to \mathbb{Z} \to \mathbb{Z}^2 \to \tilde{H}_1(X) \to 0.$$

Is it possible from this exact sequence to determine the homology groups, in some algebraic fashion? Or do I need to determine what the maps are? How do I determine the maps?

Answer 1

I have yet another solution, using cellular homology directly.

Define a CW-structure on $X$ with one 0-cell $e^0$, three 1-cells $a, b, c$, and two 2-cells $U,L$. Attach the 1-cells to $e^0$ so that $X^1 = S^1 \vee S^1 \vee S^1$, and attach the 2-cells via the relations so $\partial U \mapsto aba^{-1}b^{-1}$ and $\partial L \mapsto cac^{-1}a^{-1}$. (In the plane, this looks like a horizontally bisected rectangle with the horizontal edges & bisecting line identified in the same direction, and the opposite vertical edges identified; i.e. two flat tori glued along their top and bottom edges.) Then the nontrivial relative homology groups $H_2(X, X^1)$, $H_1(X^1, e^0)$, and $H_0(e^0)$ form the long exact sequence $$ 0 \to \mathbb Z^2 \xrightarrow{d_2} \mathbb Z^3 \xrightarrow{d_1}\mathbb Z \to 0.$$ Since there is only one 0-cell, the cellular boundary map $d_1 = 0$ (see Hatcher, paragraph before the Cellular Boundary Formula). Meanwhile one directly computes $d_2(U) = a + b - a - b = 0$ and $d_2(L) = c + a - c - a = 0$, so $d_2 = 0$. Since the sequence is exact, we get $H_2(X) = \mathbb Z^2$, $H_1(X) = \mathbb Z^3$, $H_0(X) = \mathbb Z$, and all other homology groups are $0$.

Answer 2

This solution relies on the long exact reduced homology sequence of a NDR pair (Hatcher's Theorem $2.13$), and provides different approach to the problem from @tsho's solution.

Let us call $$\underbrace{\Huge{\mathsf x} \normalsize\times S^1}_{A}~~\subset~~ \underbrace{\Huge{\propto}\normalsize\times S^1}_{B}~~\subset~~ \underbrace{\Huge{\infty}\normalsize \times S^1}_{X}$$ All three pairs $(B,A),~(X,A),~(X,B)$ are Neighborhood Deformation Retracts, as Hatcher puts it, "good pairs". Also, it is obvious that $A$ is homotopy equivalent to the circle, and $B$ to the torus. Let us write the long exact reduced homology sequence for the the good pairs $(X,A)$ and $(X,B)$ : the morphism of pair given by the inclusion $(X,A)\hookrightarrow (X,B)$ gives following commutative diagram $$\begin{array}{c}0\to &0&\to&\tilde{H}_2(X)&\to&\tilde{H}_2(X/A)&\to&\tilde{H}_1(A)&\to&\tilde{H}_1(X)&\to &\tilde{H}_1(X/A)&\to 0\\ &\downarrow&&\Vert&&\downarrow&&\downarrow&&\Vert&&\downarrow\\ 0\to& \tilde{H}_2(B)&\to&\tilde{H}_2(X)&\to&\tilde{H}_2(X/B)&\to&\tilde{H}_1(B)&\to&\tilde{H}_1(X)&\to &\tilde{H}_1(X/B)&\to 0 \end{array}$$ Now $X/A$ is the wedge sum of two pinched spheres $P$ (the space studied in the previous question), and $X/B$ is simply a pinched sphere, and it follows that $\tilde{H}_*(X/B)\simeq\tilde H_*(P)$ and $\tilde{H}_*(X/A)\simeq\tilde H_*(P)\bigoplus \tilde H_*(P)$ where the isomorphism is given by the map $(i_*^+,i_*^-)$ where $i^+$ (resp. $i^-$) are the inclusions of $P$ as the upper (resp. lower) pinched sphere in $X/A$.

Since a pinched sphere is homotopy equivalent to a sphere with a diameter attached to it, which in turn is homotopy equivalent to the wedge sum of a sphere and a circle, we have $\tilde H_*(P)\simeq \Bbb Z\oplus\Bbb Z$ concentrated in degree $1$ and $2$. We can now replace the above diagram by the following simpler one

$$\begin{array}{c}0\to &0&\to&\tilde{H}_2(X)&\to&\Bbb Z\oplus \Bbb Z &\stackrel{\gamma}{\to}&\Bbb Z &\to&\tilde{H}_1(X)&\to & \Bbb Z\oplus \Bbb Z &\to 0\\ &\downarrow&&\Vert&&~~\downarrow\beta&&\downarrow&&\Vert&&\downarrow\\ 0\to& \Bbb Z &\to&\tilde{H}_2(X)&\stackrel{\alpha}{\to}& \Bbb Z &\to& \Bbb Z\oplus \Bbb Z &\to&\tilde{H}_1(X)&\to & \Bbb Z &\to 0 \end{array}$$

From the left side of this diagram, it follows that $\tilde H_2(X)$ is a subgroup of $\Bbb Z\oplus \Bbb Z$ containing a copy of $\Bbb Z$, so $\tilde H_2(X)\simeq\Bbb Z$ or $\Bbb Z\oplus\Bbb Z$. Let us assume $\tilde H_2(X)\simeq \Bbb Z$ and try to derive a contradiction.

Since $\Bbb Z$ is torsion free, we must have $\alpha=0$. The vertical map $\beta$ is onto as it corresponds to collapsing the lower copy of $P$ inside $P\vee P\simeq X/A$ to a point, and thus $\beta$ is the projection onto the first factor. The commutativity of the diagram then forces the image of $\tilde H_2(X)$ to lie inside $\Bbb Z\oplus 0\subset \Bbb Z\oplus \Bbb Z$. However, there is an obvious self-homeomorphism of $X$ interchanging the upper and lower toruses of $X$ which passes to the quotient, and permutes the two factors $\Bbb Z \oplus \Bbb Z=\tilde H_2(X/A)$ (and possibly adds a sign). Thus, the image of $\tilde H_2(X)$ inside $\Bbb Z \oplus \Bbb Z$ is contained in $\Bbb Z\oplus 0\cap 0\oplus \Bbb Z=0$, but this contradicts the injectivity of the top left arrow. The same argument works when we replace $B$ with $B'=T(B)$ where $T$ is the self-homeomorphism of $X$ that interchanges the two circles in the wedge sum $S^1\vee S^1$. The new map $\beta'$ is the projection onto the second factor, so the map $\tilde H_2(X)\to \Bbb Z\oplus\Bbb Z$ sends $\tilde H_2(X)$ into $\ker(\beta)\cap\ker(\beta')=\Bbb Z\oplus 0\cap 0\oplus \Bbb Z=0$ contradicting injectivity.

As a consequence, $$\tilde H_2(X)\simeq \Bbb Z\oplus\Bbb Z$$

To finish the proof, we note that by the standard theory of finitely generated abelian groups, the quotient of $\Bbb Z\oplus\Bbb Z$ by a subgroup $S$ isomorphic to $\Bbb Z\oplus\Bbb Z$ is the product of two cyclic groups, and cannot be a subgroup of $\Bbb Z$ unless the subgroup $S$ is all of $\Bbb Z\oplus\Bbb Z$. This forces the top left arrow $\tilde{H}_2(X)\to\Bbb Z\oplus \Bbb Z $ to be an isomorphism, and $\gamma=0$. The top sequence then degenerates to a short exact sequence $$0\to\Bbb Z \to\tilde{H}_1(X)\to \Bbb Z\oplus \Bbb Z \to 0$$

Consequently, $$\tilde{H}_1(X)\simeq \Bbb Z\oplus\Bbb Z\oplus\Bbb Z$$

Answer 3

Firstly, I believe that $H_1(\mathrm{Torus})=\mathbb{Z}^2$ hence the sequence should be $$0 \xrightarrow 0 \mathbb{Z}^2 \xrightarrow f \tilde{H}_2(X) \xrightarrow g \mathbb{Z} \xrightarrow h \mathbb{Z}^2 \oplus \mathbb{Z}^2 \to \tilde{H}_1(X) \to 0.$$ Now, you need to determine at least one map to solve this. In Mayer Veatoris sequence, f is $(t,s) \mapsto i_*t+i_*s$, g is $t\mapsto \partial t$, and h is $t \mapsto (i_*t,-i_*t)$ where all of my $i_*$'s are the homomorphims related to the appropriate canonical inclusions. (At this point you should try solving the question yourself and check the answer for feedback).

We'll start by using the exactness of the sequence: $\operatorname{im}0=0=\operatorname{ker}f$ means $f$ is injective, $\operatorname{im}f=\operatorname{ker}g$ tells us nothing yet, and $\operatorname{im}g=\operatorname{ker}h$.

It will be most natural to determine $h$, since its the only one whose both domain and range are known. So, $h$ takes a generator of the circle, embeds it in each of the tori, and the embedded circle inside each of the tori we know to be a generator of one of his $\mathbb{Z}$'s. ($H_1(T)=\mathbb{Z}^2$, a $\mathbb{Z}$ for the "horizontal" circle and a $\mathbb{Z}$ for the "vertical" circle, in our case, the generator of the original $\mathbb{Z}$ is being mapped to the horizontal circle, that is if you draw the torus "horizontally" as you normally would, and draw $X$ as two tori ("tires") stacked on top of each other).

So we can write $h(1)=((1,0),(1,0))$ and so $\operatorname{im}g=\operatorname{ker}h=0$, hence $g=0$, so $\operatorname{im}f=\operatorname{ker}0=$ entire group, meaning $f$ is surjective and hence an isomorphism. So $H_2(X)=\mathbb{Z}^2$. $H_1$ could be obtained similarly or by using $H_1=\operatorname{Ab}(\pi _1)$.

Answer 4

I think I have a third solution which computes the homology from the cell complex directly. I realize that this is more than a year late so who knows if anyone will find this useful but I might as well add it because it might be useful to someone else looking for solutions.

Let $X$ be $S^{1} \times (S^{1} \vee S^{1}).$ Then, we can embed the torus $S^{1} \times S^{1} = T^{2}$ inside $X$ in such a manner that $X$ retracts onto the torus (just send both tori to the same torus without messing with the glued up diameter).

So, because retracts cause exact sequences to split, we have

$$\tilde{H}_{*}(X) = \tilde{H}_{*}(X/T^{2}) \oplus \tilde{H}_{*}(T^{2}).$$

Now, $X/T^{2}$ is a pinched torus (a torus with one circle squished to a point). This has the following cell decomposition:

1 0-Cell p

1 1-Cell (the other $1$ cell degenerates)

1 2-Cell

As in the case of the cellular boundary maps on the torus, all boundary maps are trivial (the boundary maps are the same after setting the degenerate $1$ cell to 0). Hence,

$$\tilde{H}_{*}(X) = (0, \mathbb{Z} \oplus \mathbb{Z}^{2}, \mathbb{Z} \oplus \mathbb{Z}).$$

Answer 5

If your coefficients are in a field, then you can use the Künneth formula, which says that if $X=Y \times Z$ $$ H^i(X) = \bigoplus_{j+l=i} H^j(Y) \otimes H^{l}(Z). $$

So, for example to calculate $H^1(X)$, there are two terms in sequence, namely when $(j,l)=(0,1)$ and when $(j,l)=(1,0)$. So $$ H^1(X) = H^1(Y) \otimes H^0(Z) \bigoplus H^0(Y) \otimes H^1(Z).$$

In your case, $Y=S^1$ and $Z=S^1 \lor S^1$, so that $$ H^1(X) = H^1(S^1) \otimes H^0(S^1 \lor S^1) \bigoplus H^0(S^1) \otimes H^1(S^1 \lor S^1)$$ $$ = k \otimes k \bigoplus k \otimes(k^2) = k \oplus k^2 = k^3.$$

Answer 6

Using a suitable Mayer-Vietoris sequence we have:

If $X$ is a nonempty space then: $$H_n(X\times S^1)\cong H_n(X)\oplus H_{n-1}(X),\quad\quad n\ge1$$

Since $X:=S^1\vee S^1$ is well known to have homology $\Bbb Z,\Bbb Z^2,0,0,\cdots$ it's immediate that $H_1(S^1\times(S^1\vee S^1))\cong\Bbb Z^2\oplus\Bbb Z\cong\Bbb Z^3$ and $H_2(S^1\times(S^1\vee S^1))\cong 0\oplus\Bbb Z^2\cong\Bbb Z^2$ and all higher homology is trivial.