I am reviewing old algebraic topology qualifying exams, and I need to compute the fundamental group and the homology of the following space.
First, a brief calculation of $\pi_1(X)$. Let $C$ be the sphere and let $D$ be the torus unioned with some small path connected open neighborhood of the intersection of $C$ and $D$. Note that $C\cap D$ (path-connected) will look like an oval on the sphere with two holes in it. It will deform rectract to $S^1\vee S^1$. Thus we have that $\pi_1(C)$ is trivial, $\pi_1(D)=\langle a,b| a b a^{-1} b^{-1}\rangle$ and $\pi_1(C\cap D)=\langle c,d| \hspace{3mm}\rangle$. We then have $(i_C)_*(c)=(i_C)_*(d)=0$ where $i_C:C\cap D\hookrightarrow C\cong S^2$ is the inclusion. $(i_D)_*(d)=(i_D)_*(d)=a\in \pi_1(T^2)$ where $i_D:C\cap D\hookrightarrow D\cong T^2$ is the inclusion. So SVK's theorem gives
$$ \pi_1(X)= \langle a, b | a b a^{-1} b^{-1}, a\rangle \cong \mathbb{Z}$$
Next, I need to calculate the homology of $X$. My first idea is to use the Mayer-Vietoris long exact sequence. I know that $H_0(X)$ is just $\mathbb{Z}$ (one path component) and that $H_1(X)=\pi_1^{ab}(X)\cong \mathbb{Z}$, but I would like to calculate $H_1(X)$ and $H_2(X)$ by hand for practice. Using $A\cong S^2$ and $B\cong T^2$ (perphaps with some small retractible neighborhoods so that $X=\textrm{int}(A)\cup \textrm{int} (B))$.
$(\Psi_1)_*$ will map $[a]=[(1,0)]\in \pi_1(T^2)$ to $[0]\in H_1(X)$ because that loop on the torus is homotopic to a constant map when it is seen as a loop in $S^2$. $(\Psi_1)_*$ will map $[b]=[(0,1)]\in \pi_1(T^2)$ to $[1]\in H_1(X)$. However, this analysis is based on knowledge of $\pi_1(X)$, which is isomorphic to $H_1(X)$, so this kind of reasoning seems contingent on the fact that I already know that $H_1(X)=\langle b| \hspace{3mm}\rangle$. I think that that $(\Phi_1)_*$
will be neither injective nor surjective. It does seem to split and map one copy of $\mathbb{Z}$ to $0$ and the other copy of $\mathbb{Z}$ to the second summoned of $H_1(T^2)=\mathbb{Z}\oplus \mathbb{Z}$. $\textrm{Im}(\Phi_1)_*\cong\mathbb{Z}\cong\textrm{Ker} (\Psi_1)_*$ by exactness. So I'm unsure how to compute $H_1(X)$ without relying on the fundamental group.
for $n=2$ we have
It is immediate that $(\Phi_2)_*$ is injective. I know that given $[z]\in H_2(X)$ ($[z]$ is a cycle) , I can write $z=x+y$ with $x,y\in C_2(A)\oplus C_2(B)$, then $0=\partial([z])=\partial(z)+\partial(y)$ and $\delta_2([z])=[\partial(x)]=-[\partial(y)]$. I haven't decided on a cell decomposition for $X$, so I feel like I cannot proceed any further with the analysis of these maps. Is there a straight-forward way of proceeding with my long exact sequence, or would I be better off looking for a cell decomposition and computing the homology that way?
Best Answer
As mentioned by JHF, $X$ is actually $S^1 \vee S^1 \vee S^2 \vee S^2 \vee S^2$. To see that, take a look at #1 here: https://www.math3ma.com/blog/clever-homotopy-equivalences, and contract what you called the oval, so that both halves of the torus become spheres with two identified points.
Your approach does lead to a result as well, but you have to be more careful. First of all, I don't think that $\pi_1(D)=\langle a,b| a b a^{-1} b^{-1}\rangle$. (Your definition of $D$ is recursive but I think I got it.) $D$ is more than a torus, and, accordingly, you seem to have lost one generator: "from the side" a torus looks like a big circle whereas $D$ looks like a wedge of two circles due to the component you've added.
Next, I don't think that to say what you've said about $(\Psi_1)_*$ is to use $\pi_1(X)$. One generator is mapped to zero and the image of the other generates a free subgroup - that's all you need to say here, no $\pi_1$ references. Then you can follow through with this calculation: first, by looking at what happens in $H_0,$ you can find the image of $\delta_1$ (as the kernel of $(\Phi_0)_*$). Then you can also find its kernel as the image of $(\Psi_1)_*$. Both will be $\mathbb{Z}$, which yields the right answer.
$H_2$ is even simpler. $(\Phi_2)_*$ is injective, of course, but you can say more: it's obviously zero. So $(\Psi_2)_*$ is injective. Thus you have a $\mathbb{Z}\oplus \mathbb{Z}$ subgroup in $H_2(X)$ and the quotient over it is the image of $\delta_2$, which is the kernel of $(\Phi_1)_*$, which is $\mathbb{Z}$. So, $H_2(X)=\mathbb{Z}^3.$ (Both times I'm using that extensions with the quotient a free group are trivial.)