Theorem
If
is a continuous,, open function from a locally compact space
onto a space
then
is also locally compact.
Proof
A function
is said to be open if, for any open subset![]()
is open in![]()
Let
and let
be a neighbourhood of
For some![]()

Since
is continuous, an open set
exists such that
and![]()
Since
is locally compact, there is a compact set
such that![]()
Then![]()
Since
is open,
is open. Since
is compact and
is continuous,
is compact. Hence
is locally compact.