Question

Suppose that p: X + X is a covering map. Recall the definition of Deck(p) and prove it is a group (using composition of functions as the binary operation).

Answer 1

For notational simplification we'll assume C as a covering space instead of X tilde.

Recall a surjective continuous map
p:0 X
, is said to be a covering map if for every
re X
, there exists an open set
IEUX
, such that
p-1(U)
is a disjoint union of open sets of C each of which are homeomorphic to U via p.

Suppose
p:0 X
be a covering map. Then an homeomorphism
1:0 +
, is said to be a deck transformation if
p=pot
.

To show is a group.

Deck(p)

Let $f_1$ and $f_2$ be two elements in
Deck(p)
, to show
fi o f2 e Deck (p)
.

Note that compoaition of two homeomorphics are again a homeomorphism and $p\circ (f_1\circ f_2)=(p\circ f_1)\circ f_2=p\circ f_2=p$ , hence is in .

Deck(p)

As composition respect associative law this operation as again associative.

Note that the indentity map plays the role of group identity as $i:C\to C$ is a homeomorphism which satisfies the condition $p\circ i=p$.

For any
fe Deck (p)
inverse will be $f^{-1}$, as f being a homeomorphism $f^{-1}$ is also a homeomorphism, and $f\circ f^{-1}= f^{-1} \circ f=i$ , and $p\circ f=p=> p=p\circ f\circ f^{-1}=p \circ f^{-1}$ , hence a group.

Feel free to comment if you have any doubts. Cheers!