Remark 3

Let

$X_1^{g_1}$ X $X_2^{g_2}$ X

(15) $p_{_1}p$ and (16) $p_{_2}p$

$B_1^{f_1}$ B $B_2^{f_2}$ B

be cartesian squares. Consider the square

$X_1\times_XX_2^g$ X

$B_1\times_BB_2^f$ B where $f$ (resp. $g$) is the mapping which defines the structure of B-space (resp. of X-space) of the fibred product $B_1\times_BB_2$ (resp. of the fibred product $X_1\times_XX_2$), and where $p'$ is the mapping deduced from $p_1\times p_2$ by passing to subsets. It is cartesian (I, p. 13, cor. 2 of prop. 5).

Consider then the following two commutative squares:

and

Square (17) is composed of squares (15) and (18), as well as of squares (16) and (19). By prop. 7, squares (18) and (19) are cartesian.