For non-reduced singular homology we know that the inclusions $\imath_i : X_i \hookrightarrow \bigsqcup_{i \in I} X_i$ to a disjoint union induce an isomorphism $\oplus_{i \in I} (\imath_i)_{\ast} : \bigoplus_{i \in I} H_n(X_i) \longrightarrow H_n( \bigsqcup_{i \in I} X_i )$ for any $n \geq 0$. Is it true that there is also an isomorphism on homology groups $\oplus_{i \in I} (\imath_i)_{\ast} : \bigoplus_{i \in I} \tilde{H}_n(X_i) \longrightarrow \tilde{H}_n( \bigsqcup_{i \in I} X_i )$?
At first I thought there should be such an isomorphism, but let $X = \mathbb{S}^1 \sqcup \mathbb{S}^1 $. Then, $H_0(X)= H_0(\mathbb{S}^1) \oplus H_0(\mathbb{S}^1) = \mathbb{Z} \oplus \mathbb{Z}$ and $\tilde{H}_0(X)=H_0(X)/\mathbb{Z} = \mathbb{Z}$. But on the other hand if such an isomorphism existed, $\tilde{H}_0(X) = \tilde{H}_0(\mathbb{S}^1) \oplus \tilde{H}_0(\mathbb{S}^1) = 0 \oplus 0 = 0$ and i get a contradictory result.
So the question is, why does this isomorphism fail? Why isn't the result true in the reduced case?
Best Answer
It is true for $n> 0$ because $H_n(Y) =\tilde H_n(Y)$ in this case. For $n = 0$ it is never true unless $I$ has only one element.
To see this, recall that reduced homology groups are only defined for non-empty spaces. Now let $p_i : X_i \to *$ denote the unique map into a one-point space $*$ and $j_i : * \to X_i$ be any map. Then $p_i \circ j_i = id$. Let $D = \bigsqcup _{i \in I} *$. The maps $p_i$ and $j_i$ induce maps $p : X = \bigsqcup _{i \in I}X_i \to D$ and $j : D \to \bigsqcup _{i \in I}X_i$ such that $p \circ j = id$. We have $p_* \circ j_* = id$ on $\tilde H_0$, thus $p_* : \tilde H_0(X) \to \tilde H_0(D)$ is onto.
Consider the following commutative diagram:
$\require{AMScd}$
$$\begin{CD} \bigoplus_{i \in I}\tilde H_0(X_i) @>\bigoplus_{i \in I} (\imath_i)_{\ast}>> \tilde H_0(X)\\ @V\bigoplus_{i \in I} (p_i)_{\ast}VV@VVp_*V\\ \bigoplus_{i \in I}\tilde H_0(*) @>\bigoplus_{i \in I} (\imath_i)_{\ast}>> \tilde H_0(D)\\ \end{CD}$$ We have $\bigoplus_{i \in I}\tilde H_0(*) = 0$, but $\tilde H_0(D) \ne 0$ if $I$ has more than one element. If $\bigoplus_{i \in I} (\imath_i)_{\ast} : \bigoplus_{i \in I} \tilde{H}_n(X_i) \longrightarrow \tilde{H}_n( X)$ were an isomorphism (for any family of $X_i$), then $\bigoplus_{i \in I} (\imath_i)_{\ast} : \bigoplus_{i \in I} \tilde{H}_n(*) \longrightarrow \tilde{H}_n(D)$ would be onto - but this is impossible.