You should probably start off with the definition of a homo-morphism (or an iso-morphism for bijective spaces) and prove the group axioms one by one.
I hope you are well.
I need your help:
Suppose X and Y are topological homeomorphic spaces such that X is a group. Then is it true that Y is also a group?
Two spaces are said to be homeomorphic if there is a bijective continuous function between them and the inverse of the function is also continuous.
Group is algebraic structure. a set X with binary operation * is a group if (1) * is associative that is x*(y*z)=(x*y)*z for all x,y,z in X. (2) there is an element e in X called the identity element which satisfy e*x=x*e=x, for all x in X. (3) for each element x in X there is an inverse x^-1 such that x*x^-1=x^-1*x=e.
Please help me and every guidance is highly appreciated.
Thank you in advance
Suppose X = Y = , with the usual topology on both, and our homeomorphism is the identity map. If the operation in Y is multiplication, and the operation in X is addition, then even though X and Y are homeomorphic as topological spaces, they aren't algebraically isomorphic, because Y isn't even a group (0 has no inverse).
Another question: Given two homeomorphic spaces, each a group with a binary operator, does there exist a homeomorphism which is also an isomorphism? That is a more complicated question. The answer is no in general. Let G and H be two non-isomorphic groups of equal cardinality. Give each the discrete topology. Then, any bijection between the two is a homeomorphism (since every function between two spaces with discrete topologies will be continuous). But, since you selected the groups to be non-isomorphic, there is no isomorphism between them.
Previous reply was within context of the OP question.
Group: *, associativity,identity,inverse.
Homeomorphism: defined between groups and (a*b)’=a’*’b’
Y is trivially a group by definition (Homeomorphism defined between groups).
Homeomorphism is not a homomorphism. Definition is correct.
Let f:X →Y be a homeomorphism and let X be a group under *.
Let x’=f(x), y’=f(y), …..
Let x’*’y’=z’ where z=x*y. Then Y is a group under *’.
x’*’y’=e’, e=x*y inverse
Example f(x) = x’ = x2 and * = +.
Then x’*’y’=z’ → x1/2*’y1/2=(x+y)1/2, which is a result of the definition of *’.
f, *,*’ is not necessarily a homomorphism between X and Y.