Corollary

Let

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

$p'p$

${B'}^f$ B

be a Cartesian square.

a) For every point $b'$ of $B'$, the mapping $f'$ induces a homeomorphism of the fibre $X'_{b'}$ of $p'$ onto the fibre $X_{f(b')}$ of $p$.

b) If the mapping $p$ is injective (resp. surjective, resp. bijective), the same is true of $p'$.

Let $b'$ be a point of $B'$. Put $b=f(b')$. For every $x'\in X'_{b'}$, one has $p(f'(x')) =f(p'(x')) =f(b') =b$, whence $f'(x')\in X_b$. This proves that

one has $X'_{b'}\subset (\overset{-1}{f}')(X_b)$. In Prop. 4, take $B_0=\{b\},B'_0=\{b'\}$ and $X_0= X_b$; one then has $X'_0= X'_{b'}$, whence assertion a).

For the mapping $p$ to be injective (resp. surjective, resp. bijective), it is necessary and sufficient that the cardinal of each of its fibres be less than (resp. greater than, resp. equal to) 1. Assertion b) follows.