Let $X_1$ and $X_2$ be topological spaces. Let's denote by $\pi_0(X)$ the set of path components of $X$. I would like to know if there is a relation between $\pi_0(X_1)$, $\pi_0(X_2)$, and $\pi_0(X_1\times X_2)$.
I have already showed that $\pi_0(X_1)\times \pi_0(X_2)\subset \pi_0(X_1\times X_2)$. Is the other inclusion true?
Thanks!
First thing to note is that this isn't technically a set inclusion, rather a natural inclusion $(X_i,Y_j) \mapsto X_i \times Y_j$, since the product of path connected spaces is path connected. If $X=A \cup B$, with $A \cap B = \emptyset$, then for any set $Y$, $X \times Y = A \times Y \cup B \times Y$ with this being disoint component. So if $Y = \bigcup_i Y_i$ and $X=\bigcup_j X_j$, where $X_j,Y_i$ are the path components (which are disjoint!), we have that $X \times Y = \bigcup_{ij} X_j \times Y_i$. These are clearly the path components of $X \times Y$, since if theres is path linking points in different ones there would be a path connecting $Y_i$ to $Y_i'$ or $X_j$ to $X_j'$. So we get a natural bijection (the inclusion is surjective) $\pi_0(X)\times \pi_0(Y) \rightarrow \pi_0(X \times Y)$ where $(X_j,Y_i) \mapsto X_j \times Y_i$.