I am trying to solve the following question:
Given a topological space X, show that $H_{i}(X \times S^1) = H_{i}(X) \times H_{i-1}(X)$ for $i>0$ and $H_{0}(X \times S^1)=H_0(X)$.
The (almost) same problem was posed here: Calculating Homology Groups of $S^1\times X$, however there is no answer and it uses the additional assumption that the homology groups of X are free.
I have the following questions:
- Can the assumption of freeness be left out?
- Can this be generalised to $S^n$ for larger n?
Note that I have managed to solve the problem with the assumption of freeness, using Mayer-Vietoris:
Set $A=X \times S^1 \setminus \{p\}$ and $B=X \times S^1 \setminus \{q\}$ for distinct points $p, q$ and $Y=X \times S^1$.
Then Mayer-Vietoris gives the following long exact sequence:
$\cdots \rightarrow H_p(A\cap B) \rightarrow H_p(A)\oplus H_p(B)\rightarrow H_p(Y)\rightarrow H_{p-1}(A\cap B)\rightarrow \cdots$
Denote the homomorphisms $d,i_A\oplus i_B, j_A+j_B$ in this order. Here A and B are homotopy equivalent to X and the intersection is homotopy equivalent to two copies of X.
We can use the long exact sequence above to create a short exact sequence:
$0 \rightarrow ker(d) \rightarrow H_p(Y) \rightarrow Im(d) \rightarrow 0$
By exactness we have $Im(d)=ker(i_A\oplus i_B)$ and
$ker(d)=Im(j_A+j_B)=(H_p(A)\oplus H_p(B))/ker(j_A+j_B)=(H_p(A)\oplus)H_p(B))/ker(i_A\oplus i_B)$.
So I only need to compute $i_A\oplus i_B$.
Given a cycle $c_p\in S_p(X)$, take corresponding cycles $a_p, b_p$ in the two copies of X that form $A\cap B$. Then $i_A\oplus i_B$ takes both $a_p, b_p$ to $(c_p, -c_p)$ in $H_p(A)\oplus H_p(B)$. Thus $ker(i_A\oplus i_B)=\langle a_p-b_p \rangle$ and $Im(i_A\oplus i_B)=\langle (c_p,-c_p) \rangle$.
The above short exact sequence becomes
$0 \rightarrow (H_p(A)\oplus H_p(B))/\langle (c_p,-c_p) \rangle \rightarrow H_p(Y) \rightarrow \langle a_{p-1}-b_{p-1} \rangle \rightarrow 0$
The first group in this sequence is isomorphic to $H_p(X)$ and the last one to $H_{p-1}(X)$. So, if I know that $H_{p-1}(X)$ is free, the sequence automatically splits and we are done. By the way that $d$ is defined, I would expect that the sequence splits even without the additional assumption, but I have not been able to find $H_{p-1}(X)\rightarrow H_p(Y)$ that satisfies the splitting condition.
Any help regarding the two questions/comments on my solution is much appreciated!
Best Answer
Let us understand $S^1$ as the set of complex numbers $z$ with $\lvert z \rvert = 1$.
In Example 2.48 Hatcher derives an exact sequence "which is somewhat similar to the Mayer–Vietoris sequence and which in some cases generalizes it."
Adapted to your question it is $$0 \to H_n(X) \xrightarrow{i_*} H_n(X \times S^1) \to H_{n-1}(X) \to 0 \tag{1}$$ where $i : X \to X \times S^1, i(x) = (x,1)$.
This sequence obviously splits. In fact, the projection $\pi : X \times S^1 \to X$ has the property $\pi_* \circ i_* = (\pi \circ i)_* = id_* = id$.
As you have shown in your question, this suffices to prove the desired result.
Hatcher's above-mentioned sequence is more general than $(1)$ and requires its own proof. However, the special sequence $(1)$ can be derived from the Mayer-Vietoris sequence.
As in your question, let $A = X \times (S^1 \setminus \{i\})$ and $A = X \times (S^1 \setminus \{-i\})$. The Mayer-Vietoris sequence is $$\ldots H_n(A \cap B) \xrightarrow{\phi_n} H_n(A) \oplus H_n(B) \xrightarrow{\psi_n} H_n(X \times S^1) \xrightarrow{d_n} H_{n-1}(A \cap B) \ldots $$ Here
$\phi_n(g) = ((i_A)_*(g), (i_B)_*(g))$ with inclusions $i_A : A \cap B \to A$ and $i_B : A \cap B \to B$
$\psi_n(a,b) = (j_A)_*(a) - (j_B)_*(b)$ with inclusions $j_A : A \to X \times S^1$ and $j_B : B \to X \times S^1$
This produces a short exact sequence $$0 \to (H_n(A) \oplus H_n(B))/\ker \psi_n \xrightarrow{\hat \psi_n} H_n(X \times S^1) \xrightarrow{d_n} \operatorname{im} d_n \to 0 \tag{2}$$ By exactness we have $\operatorname{im} d_n = \ker \phi_{n-1}$.
For each $M \subset S^1$ with $1 \in M$ let
$X_M = X \times M$
$i_M : X \to X_M, i_M(x) = (x,1)$
$\pi_M : X_M \to X, \pi:M(x,z) = x$.
With $S_+ = S^1 \setminus \{i\})$, $S_- = S^1 \setminus \{-i\})$ and $S = S^1 \setminus \{i,-i\})$ we get $X_{S_+} = A$, $X_{S_-} = B$, $X_S = A \cap B$. The above $i : X \to X \times S^1$ is nothing else $i_{S^1}$.
We have $\pi_M \circ i_M = id$, thus $(\pi_M)_* \circ (i_M)_* = id$ which proves 1.
Surjectivity follows from 1. We have
$$i_*(r_n(a, b)) = i_*((\pi_{S_+})_*(a)) - i_*((\pi_{S_-})_*(b)) = (i \circ \pi_{S_+})_*(a) - (i \circ \pi_{S_-})_*(b) .$$
We can write $i \circ \pi_{S_+} = id_X \circ r$ with $r : S_+ \to S^1, r(z) = 1$. Since $S_+$ is contractible, $r$ is homotopic to $j : S_+ \hookrightarrow S^1$. Hence $i \circ \pi_{S_+} \simeq id_X \circ j = j_A$, thus $(i \circ \pi_{S_+})_*(a) = (j_A)_*(a)$. Similarly $(i \circ \pi_{S_-})_*(b) = (j_B)_*(b)$.
Let us finally determine $\ker \phi_n$.
Since $X \times \{-1, 1\}$ is a strong deformation retract of $X_S = A \cap B$, we get $A \cap B \simeq X \times \{-1, 1\}$ and thus $H_n(A \cap B) \approx H_n(X) \oplus H_n(X)$. Let $i'_S : X \to X_S = X \times S, i'_S(x) = (x,-1)$. An explicit isomorphism $\beta_n : H_n(X) \oplus H_n(X) \to H_n(A \cap B)$ is given by $\beta_n(u,v) = (i_S)_*(u) + (i'_S)_*(v)$.
Define $e_n : H_n(X) \to H_n(X) \oplus H_n(X), e(w) = (w, -w)$. This is an injective homomorphism.
Let $c = \beta_n(u,v)$. Then $\phi_n(c)= 0$ iff $(i_A)_*(\beta_n(u,v)) = 0$ and $(i_B)_*(\beta_n(u,v)) = 0$. We have $(i_A)_*(\beta_n(u,v)) = (i_A)_*((i_S)_*(u))) +(i_A)_*((i'_S)_*(v))) = (i_A \circ i_S)_*(u) +(i_A \circ i'_S)_*(v)$. But $i_A \circ i'_S \simeq i_A \circ i_S = i_{S_+}$, thus $(i_A)_*(\beta_n(u,v)) = (i_{S_+})_*(u) + (i_{S_+})_*(v) = (i_{S_+})_*(u + v)$. Since $(i_{S_+})_*$ is injective, we see that $(i_A)_*(\beta_n(u,v)) = 0$ iff $u + v = 0$. Similarly $(i_B)_*(\beta_n(u,v)) = 0$ iff $u + v = 0$. This shows that $\phi_n(c)= 0$ iff $u + v = 0$, i.e. $(u,v) = e_n(u)$. Hence $\phi_n(c) = 0$ iff $c \in \operatorname{im}(\beta_n \circ e_n)$.
Using 3. and 4. we can transform sequence $(2)$ into $(1)$.
Your question whether this can be generalized to $S^n$ for $n > 1$ can also be answered in the affirmative. We get $$H_k(X \times S^n) \approx H_k(X) \oplus H_{k-n}(X) \text{ for } k \ge n ,$$ $$H_k(X \times S^n) \approx H_k(X) \text{ for } k < n .$$
The proof is based on the Künneth theorem.