Proposition 2

Let B, $B'$ and X be topological spaces and let $p: X\rightarrow B,f: B'\rightarrow B$ be continuous mappings.

a) The square

$B'\times_BX^{pr_2}$ X

(2) pr$_{_1}p$

${B'}^f$ B

is a Cartesian square.

b) For every commutative square

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

${B'}^f$ B

there exists a unique continuous mapping $h: X'\rightarrow B'\times_BX$ such that pr$_1\circ h=p'$ and pr$_2\circ h=f'$.

c) The commutative square (3) is cartesian if and only if $h$ is a homeomorphism.

Assertion a) results from Prop. 1 and from the universal property of the product B-space of two B-spaces (I, p. 3). Assertion b) follows from it.

If square (3) is cartesian, there exists a unique continuous mapping $h': B'\times_BX\rightarrow X'$ such that $f'\circ h'$ = pr$_2$ and $p'\circ h'=$ pr$_1$. We have $f'\circ h'\circ h=f'$ and $p'\circ h'\circ h=p'$, whence $h'\circ h=$ Id$_{X'}$ since square (3) is cartesian. We have pr$_1\circ h\circ h'$ = pr$_1$ and pr$_2\circ h\circ h'$ = pr$_2$, whence $h\circ h'=$ Id$_{B'\times_BX}$ since square (2) is cartesian. This proves that $h$ is a homeomorphism.

Conversely, suppose that $h$ is a homeomorphism; since square (2) is cartesian, square (3) is also cartesian.

The mapping $h: X'\rightarrow B'\times_BX$ whose existence and uniqueness is asserted by assertion b) of the preceding proposition will be said to be canonical: it is the mapping denoted by $(p', f')$ in I, p. 3.