Proof of Homeomorphism: An Example

[beetle2]

Hi Guy's
I need to show that two spaces are Homeomorphic for a given function between them.
Is there an online example of a proof.

A lot of text on the web tells you what it needs to be a homeomorphism but I not an example of a proof. I just want an good example I can you to help me.

Thanks in advance

 

[zhentil]

The definition of homeomorphism (map is continuous, as is its inverse) is also the strategy for the proof.

 

[Werg22]

What you gotten down so far? Is the domain connected? The codomain Hausdorff?

 

[beetle2]

What I've got so far is...

I must say in advance that this is an assignment question.

I have been given the following.

Let (X,T) be a topological space. Let I := [0,1]:= {t \in \Real \mid 0 \leq t \leq 1}
be endowed with the Euclidean topology. Prove that for each \lambda \in [0,1] the function:

i_{\lambda}: X \rightarrow X \times I, x \rightarrow(x,\lambda)

is a homeomorphism of X onto im(i_{\lambda}), where X \times I is endowed with the product topology.

I know that if two spaces are homeomorhic you need a function between the spaces that satisfy.


1: F is one-one
2: F is onto
3; A subset A \subset Xis open if and only if f(A) is open.



Therfore we need to show that the inverse function i_{\lambda}^{-1}(t_0 \times \sigma_{\lambda}) is open in A whenever

(t_0 \times \sigma_{\lambda}) is open in X \times I where t \in T


but (t_0 \times \sigma_{\lambda}) open implies t_0 \in T
, \sigma_{\lambda} = [\lambda - \epsilon_{1},\lambda - \epsilon_{2}} and i_{\lambda}^{-1}(t_0 \times \sigma_{\lambda}) = t_0

Since (x,\lambda) \in t_0 \times \sigma_{\lambda i})implies x \in t_0 , \lambda \in \sigma_{\lambda}

Therefore i_{\lambda} is a homeomorphism

 

[Bacle]

What I've got so far is...

I

I know that if two spaces are homeomorhic you need a function between the spaces that satisfy.


1: F is one-one
2: F is onto
3; A subset A \subset Xis open if and only if f(A) is open.


<snip>

(t_0 \times \sigma_{\lambda}) is open in X \times I where t \in T


but (t_0 \times \sigma_{\lambda}) open implies t_0 \in T
, \sigma_{\lambda} = [\lambda - \epsilon_{1},\lambda - \epsilon_{2}} and i_{\lambda}^{-1}(t_0 \times \sigma_{\lambda}) = t_0

Since (x,\lambda) \in t_0 \times \sigma_{\lambda i})implies x \in t_0 , \lambda \in \sigma_{\lambda}

Therefore i_{\lambda} is a homeomorphism

I think it may be clearer if you invert the order here: in order to show that
(X,T_X) and (Y,T_Y are homeomorphic to each other,
you must find a function f so that :

1) f is continuous

2)f^-1 is also continuous.

From these it follows that f has to be bijective. So in this case, first show continuity
of f :

1)take an open set in XxI product ( or take a basic or subbasic open set, easier)

and show its inverse image is open in X . Then show that f^-1 is also

continuous; like you said, this implies that if you take any O open in X , then

f^-1(O) must be open in the product space XxI


HTH.