Proposition 5

Let I be a set and, for every $i\in I$, let

be a Cartesian square. The square

(where $f,f',p,p'$ are the extensions of the families $(f_i)$, $(f'_i)$, $(p_i)$, $(p'_i)$ to the products) is Cartesian.

Let Y be a topological space, let $u: Y\rightarrow \prod_iB'_i$ and $v: Y\rightarrow \prod_iX_i$ be continuous mappings such that $f\circ u=p\circ v$. For $i\in I$, put $u_i$ = pr$_i\circ u$ and $v_i=$ pr$_i\circ v$; one has $f_i\circ u_i=p_i\circ v_i$ and there exists a unique continuous mapping $w_i: Y\rightarrow X'_i$ such that $p'_i\circ w_i=u_i$ and $f'_i\circ w_i=v_i$. Then the mapping $w= (w_i)$ is a continuous mapping of Y into $\prod_iX'_i$ such that $p'\circ w=u$ and $f'\circ w=v$, and it is the only one having these properties.