Proposition 6

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

${X'_i}^{f'_i}$ X

${B'_i}^{f_i}$ B

be a cartesian square. Let $X'$ and $B'$ be the sum spaces of the families $(X'_i)$ and $(B'_i)$ respectively. Let $f: B'\rightarrow B,f': X'\rightarrow X$ and $p': X'\rightarrow B'$ be the mappings deduced from the families $(f_i)$, $(f'_i)$ and $(p'_i)$ respectively. The square

${X'}^{f'}$ X

${B'}^f$ B

is cartesian.

The mappings $f,f'$ and $p'$ are continuous. The commutativity of square ($10'$) follows from its definition. Let us denote by $h_i$ the canonical homeomorphism of $X'_i$ onto $B'_i\times_BX$ (I, p. 8, prop. 2) and by $h: X'\rightarrow B'\times_BX$ the canonical mapping (loc. cit.). We have $h=h''\circ h'$ where $h': X'\rightarrow \coprod(B'_i\times_BX)$ is the homeomorphism deduced from the $h_i$ and $h'':\coprod(B'_i\times_BX)\rightarrow (\coprod B'_i)\times_BX$ is that defined in example 5 of I, p. 4. We conclude by prop. 2 of I, p. 8.