Proposition 4

Every covering of a nonempty connected space has a degree.

Let G be a topological space, let $(F, g)$ be a covering of G and let $(E, f)$ be a covering of F; suppose that these coverings have a degree. If the G-space $(E, g\circ f)$ is a covering, then it has a degree and one has deg(E$, g\circ f$) $=$ deg(F$, g$) deg(E$, f$), which can also be written

In fact, if $z$ is a point of G, all the fibres of the mapping

have for cardinal deg(E$, f$), and $\overset{-1}{g}(z)$ has for cardinal deg(F$, g$). The assertion therefore follows from the shepherds' principle (E, III, p. 41, prop. 9).

Let B and $B'$ be topological spaces, let $(E, p)$ be a covering of B and let $(E', p')$ be a covering of $B'$. Suppose that these coverings have a degree; then the covering $(E\times E', p\times p')$ of $B\times B'$ (I, p. 71, prop. 2) has a degree and one has:

deg(E $\times E', p\times p'$) $=$ deg(E$, p$) deg(E$', p'$).

In fact, for every pair $(b, b')\in B\times B'$, the fibre $(E\times E')_{(b,b')}$ is the product $E_b\times E'_{b'}$.

Let B and $B'$ be topological spaces, $(E, p)$ a covering of B, $(E', p')$ a covering of $B'$ and let

${E'}^{f'}$ E

$p'p$

${B'}^f$ B

a Cartesian square. If $B'$ is non-empty and if the covering $(E, p)$ has a degree, then the covering $(E', p')$ has a degree, equal to that of E, after I, p. 10, cor. of prop. 4.