Proposition 7

Let

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

(12) $p''p'$ and (13) $p'p$

${B''}^gB'{B'}^f$ B

be commutative squares; consider the square

${X''}^{f'\circ g'}$ X

${B''}^{f\circ g}$ B .

It is commutative. If squares (12) and (13) are cartesian, the same is true of square (14). If squares (13) and (14) are cartesian, the same is true of square (12).

Square (14) is commutative, for we have $p\circ f'\circ g'=f\circ p'\circ g'=f\circ g\circ p''$.

Let us denote by $h': X''\rightarrow B''\times_{B'}X',h: X'\rightarrow B'\times_BX$ and $h'': X''\rightarrow B''\times_BX$ the canonical continuous mappings deduced from the commutative squares (12), (13) and (14). Moreover, let us denote

the continuous mapping which associates to $(b'', x)$ the element $(b'', g(b''), x)$. It is a homeomorphism and one has $j\circ h''=$ (Id$_{B''}\times_Bh$)$\circ h'$.

Suppose that square (13) is cartesian. Then $h$ is a homeomorphism (I, p. 8, prop 2), hence the mapping Id$_{B''}\times_Bh$ is a homeomorphism, and $h''$ is a homeomorphism if and only if $h'$ is one. This means that square (12) is cartesian if and only if square (14) is cartesian (loc. cit.).

With the notation of proposition 7, one sometimes says that square (14) is the square composed of squares (13) and (12). The first assertion expresses that the square composed of two cartesian squares is cartesian. In particular, the $B''$-spaces $g^*(f^*(X))$ and $(f\circ g)^*(X)$ are isomorphic.