relation between germ space and generalized germ space
Proof. Indeed, assume that for some . Let and be a representatives of and respectively. It follows, that there exists an open neighbourhood of such that
In particular in , which completes the proof.
Proof. Assume that for some . Since is regular (because it is normal) then there exists an open neighbourhood such that the closure . Now since is normal and is a normal absolute retract, then can be extended to entire (by the generalized Tietze extension theorem). It is easily seen that any such extension gives the same element in (and it is independent on the choice of the representative ) and if is an extension of , then
because . This completes the proof.
Remark. If in addition is a topological ring (for example ), then it can be easily checked that preserves ring structures. In particular if is normal and or , then is an isomorphism of rings. Also it is a good question whether the assumptions in proposition 2 can be weakened.
|Title||relation between germ space and generalized germ space|
|Date of creation||2013-03-22 19:18:23|
|Last modified on||2013-03-22 19:18:23|
|Last modified by||joking (16130)|