Remark 2

Let $p: X\rightarrow B$ and $f: B'\rightarrow B$ be continuous mappings. The mapping $g: B'\rightarrow B'\times B$ defined by $g(b') = (b', f(b'))$ is a homeomorphism of $B'$ onto the graph G of the mapping $f$ (TG, I, p. 25, cor. 2), and the fibred product $B'\times_BX$ is identified (I, p. 6) with the subspace of $B'\times X$, inverse image of G under the mapping Id$_{B'}\times p: B'\times X\rightarrow B'\times B$. By the example of I, p. 10 and cor. 1 of prop. 5 (I, p. 11), the squares

$B'\times_BX^iB'\times XB'\times X^{pr_2}$ X

pr$_1$ Id$_{B'}\times p$ and Id$_{B'}\times pp$

${B'}^gB'\times BB'\times B^{pr_2}$ B

(where $i$ denotes the canonical injection) are cartesian, and the cartesian square

$B'\times_BX^{pr_2}$ X

pr$_1p$

${B'}^f$ B

is their composite. In other words, every cartesian square is identified with the square composed of a cartesian square obtained by product (I, p. 11, cor. 1 of prop. 5, square (8)) and of a cartesian square obtained by passing to subspaces (I, p. 10, example, square (5)).