Example 2

When I is not empty, one deduces from the mapping $(b, x)\mapsto x$ of $B\times \prod_{i\in I}X_i$ into $\prod_{i\in I}X_i$, by passing to subspaces, a homeomorphism of $\prod_BX_i$ onto the subspace of $\prod_{i\in I}X_i$ formed by the families $(x_i)_{i\in I}$ such that $p_i(x_i) =p_j(x_j)$ for all $i, j\in I$. This subspace will be called, by abuse, the fibred product of the family $(X_i)_{i\in I}$.