Singular homology of the sphere

Question

By Mayer-Vietoris, to compute the singular homology of the $S^n$, essentially we want to look at the following exact sequence

$0 \to H_1(S^1) \xrightarrow{\partial^{\ast}} H_0(S^0) \xrightarrow{k_{\ast}} H_0(B^1) \oplus H_0(B^1) \xrightarrow{j_{\ast}} \ldots$

where $k_{\ast}([a]) \to [(a,-a)]$, how to compute the kernel of $k_{\ast}$?

My reasoning is as follows, the above exact sequence is induced by

$0 \to S_1(S^1) \xrightarrow{\partial^{\ast}} S_0(S^0) \xrightarrow{k_{\ast}} S_0(B^1) \oplus S_0(B^1) \xrightarrow{j_{\ast}} \ldots$, and $[(a,-a)] = 0$ iff $(a,-a) \in im k_{\ast}$, which is true for every element, then $kerk_{\ast} = H_{0}(S^0) = \mathbb Z \times \mathbb Z$, so $H_1(S^1) = \mathbb Z \times \mathbb Z$, what is wrong with this argument? Since the same argument applies to $H_1(S^n)$ which is not equal to $\mathbb Z \times \mathbb Z$ for sure.

Answer 1

Your middle morphism ($k_\star$) is not quite right. The correct morphisms are: $$ 0 \to H_1(S^1) \to \mathbb Z \oplus \mathbb Z \overset{\begin{bmatrix} a \\ b\end{bmatrix} \mapsto \begin{bmatrix} a + b \\ - a - b\end{bmatrix}}\longrightarrow \mathbb Z \oplus \mathbb Z \overset{\begin{bmatrix} c \\ d\end{bmatrix} \mapsto c + d}{\longrightarrow}\mathbb Z \to 0$$ So the kernel of $k_\star$ has a single generator, namely, $(1,-1)$, hence $H_1(S^1) \cong \mathbb Z$.

But how can we be sure that $k_\star$ is as I've written it? I'll explain...

• Suppose $U$ and $V$ are your two open arcs covering your $S^1$, i.e. $U$ and $V$ are your two $B^1$'s. Then $U \cap V$ has two path-connected components, so $$H_0(U \cap V) \cong \mathbb Z \oplus \mathbb Z,$$ as you correctly identified. (You're also right to observe that $U \cap V$ is homotopy-equivalent to $S^0$.) Anyway, the generator $(1,0) \in H_0(U \cap V) $ is the homology class of any point in the first connected component of $U \cap V$, and the generator $(0,1) \in H_0(U \cap V) $ is the homology class of any point in the second connected component.

• $U$ has a single path-component, so we simply have $$H_0(U) \cong \mathbb Z.$$ We'll denote the generator of $H_0(U)$ as $1 \in \mathbb Z$. This generator is the homology class of any point in $U$.

• Under the inclusion map $U \cap V \hookrightarrow U$, a point in the first path-component of $U \cap V$ simply becomes a point in $U$. A point in the second path-component of $U \cap V$ also becomes a point in $U$. Therefore, the inclusion map $U \cap V \hookrightarrow U$ includes the map $$ H_0(U \cap V) \to H_0(U)$$ given by $$ \begin{bmatrix} 1 \\ 0\end{bmatrix} \mapsto 1, \ \ \ \ \ \ \ \ \ \ \ \begin{bmatrix} 0 \\ 1\end{bmatrix} \mapsto 1..$$

• The same remarks apply to $V$: the inclusion map $U \cap V \hookrightarrow V$, includes the map $$ H_0(U \cap V) \to H_0(V)$$ given by $$ \begin{bmatrix} 1 \\ 0\end{bmatrix} \mapsto 1, \ \ \ \ \ \ \ \ \ \ \ \begin{bmatrix} 0 \\ 1\end{bmatrix} \mapsto 1.$$

• By the definition in your textbook, the map $$k_\star : H_0(U \cap V) \to H_0(U) \oplus H_0(V)$$ is obtained by taking the direct sum of the two inclusion-induced maps that I described above in the previous two bullet points. There is one caveat, which is that we must insert a minus sign into the action of the second map (the one involving $H_0(V)$); this extra minus sign is a part of how $k_\star$ is defined in the textbook. So the action of $k_\star$ on our generators is: $$ \begin{bmatrix} 1 \\ 0\end{bmatrix} \mapsto \begin{bmatrix} 1\\ - 1\end{bmatrix} , \ \ \ \ \ \ \ \ \ \ \ \begin{bmatrix} 0\\ 1\end{bmatrix} \mapsto \begin{bmatrix} 1\\ - 1\end{bmatrix} , \\ $$ which matches what I wrote originally.

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For completeness, I'll also explain how to work out the action of $j_\star$.

• The full space $S^1 = U \cup V$ has a single path-component, so $$ H_0(S^1) \cong \mathbb Z$$ The generator $1 \in H_0(S^1)$ is the homology class of any point in $S^1$.

• Under the inclusion $U \hookrightarrow S^1$, a point in $U$ simply becomes a point in $S^1$. So the map $$ H_0(U) \to H_0(S^1) $$ induced by this inclusion map is given by $$ 1 \mapsto 1.$$

• Similarly, the inclusion map $V \hookrightarrow S^1$ induces the map $$ H_0(V) \to H_0(S^1) $$ defined by $$ 1 \mapsto 1.$$

• By definition, the map $$ j_\star : H_0(U) \oplus H_0(V) \to H_0(S^1)$$ is given by applying the two inclusion-induced maps described in my previous two bullet-points to $H_0(U)$ and $H_0(V)$ separately, then adding up the two answers. In other words, it is given by $$ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \mapsto 1, \ \ \ \ \ \ \ \ \ \begin{bmatrix} 0 \\ 1 \end{bmatrix} \mapsto 1.$$

At this point, it is perhaps worth checking that ${\rm Im}(k_\star) = {\rm Ker}(j_\star)$ - and this is true, because both ${\rm Im}(k_\star)$ and ${\rm Ker}(j_\star)$ are generated by $(1, -1) \in H_0(U) \oplus H_0(V)$. We should also check that ${\rm Im}(j_\star) = H_0(S^1) $, and indeed this is true as well.