1- Let $I$ be a finite set and for every $i \in I$, let $S_i$ be an arbitrary subspace of $[0,1]$ (endowed with its usual topology).
Moreover, let $S := \prod_{i \in I} S_i$ (endowed with product topology).
Does for every $n > 0$, $H_n(S) \simeq \{0\}$ (where $H_n(S)$ is the $n$-th singular homology group of $S$) ? If not, do we have sufficient conditions on $(S_i)_{i \in I}$ to obtain that $H_n(S) \simeq \{0\}$ ?
2- More generally, let $I$ be a set and for every $i \in I$, let $X_i$ be an arbitrary topological space.
Moreover, let $X := \prod_{i \in I} X_i$ (endowed with product topology).
How do we determine, for every $n > 0$, $H_n(X)$ ?
Thank you !
Best Answer
The path components of any subset $S_i \subset I$ are intervals (which may be degenerate to points), in particular they are contractible.
The path components $P_\alpha$ of $S$ are the products of the path components of the $S_i$, hence contractible. Since $$H_n(S) = \sum_\alpha H_n(P_\alpha) ,$$ we get $H_n(S) = 0$ for $n > 0$ and $H_0(S) = \sum_\alpha \mathbb Z$ which is a free abelian group whose rank equals the cardinality of the sets of path components of $S$.