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 $A \subset X$is 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 $A \subset X$is 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.