Corollary 1

Suppose that the mapping $f$ is universally strict and surjective and that the mapping $p'$ is universally strict. Then the mapping $p$ is universally strict.

Let

Y $^{g'}$ X

$qp$

C $^g$ B

be a cartesian square; it is a question of proving that the mapping $q$ is strict. By remark 3 of I, p. 16, the squares

$rp'$

and

$X'\times_XY^{pr_2}$ Y

$B'\times_BC^{pr_2}$ C

are cartesian, where $r: X'\times_XY\rightarrow B'\times_BC$ denotes the mapping induced by $(p', q)$. Since the mapping $f$ is universally strict and surjective, the same is true of the mapping pr$_2: B'\times_BC\rightarrow C$ (I, p. 20, def. 6 and I, p. 10, cor. of prop. 4). On the other hand, the mapping $r$ is strict, since $p'$ is assumed universally strict. By prop. 11, c) applied to the cartesian square (21), the mapping $q$ is strict.