Prove that a function from [0,1] to [0,1] is a homeomorphism

[docnet]

Homework Statement

prove that a function from [0,1] to [0,1] is a homeomorphism

Relevant Equations

prove that a function from [0,1] to [0,1] is a homeomorphism

let $X=\{0,p1,p_2,...,p_n,1\}$ and $Y=\{0,p1,p_2,...,p_n,1\}$ be sets equipped with the discrete topology.

for each $q_i$ in $Y$, the inverse image $h^{-1}(q_i)=p_i$ is open in $X$ w.r.t. to the discrete topology, so h is continuous.

every element y in Y has a preimage x in X, so h is onto.

every element y in Y has a unique preimage x in X. so $h$ is one-to one.

h is continuous, onto, and one to one so it is a homeomorphism from X to Y.

(unsure about this step)
in the limit as n goes to infinity, $X=Y=[0,1]$ are sets equipped with the finest topology on [0,1] and all the previous claims hold. so $h$ is a homeomorphism from [0,1] to [0,1].

 

[FactChecker]

I think that you are proving the wrong thing. The problem asks for you to prove that such a function,h, exists. You should either construct such a function or you should use some existence theorem to show that there is such a function.

 

[docnet]

I think that you are proving the wrong thing. The problem asks for you to prove that such a function,h, exists. You should either construct such a function or you should use some existence theorem to show that there is such a function.

okay, i think I was supposed to construct a quotient space that is equal to [0,1] and show that h is continuous on [0,1] ?

 

[fresh_42]

I would construct an example for $n=1$. Say $p_1\leq q_1.$ Then you can linearly stretch $[0,p_1]$ to $[0,q_1]$ and linearly compress $[p_1,1]$ to $[q_1,1].$ Since both functions have the same value $y=q_1$ at $x=p_1$ you can concatenate them. The rest goes on by induction.

 

[FactChecker]

okay, i think I was supposed to construct a quotient space that is equal to [0,1] and show that h is continuous on [0,1] ?

It says to "Prove ... there is a homeomorphism, h". So you need to either construct h or use some other existence theorem to show that h exists.

 

[docnet]

Let $h(t)=q_{i−1}+(\frac{q_i−q_{i−1}}{p_i−p_{i−1}})(t−p_{i−1})$, a linear mapping between the intervals $[p_i, p_{i+1}]$

required conditions: $h(0)=0, h(1)=1$, and $h(p_i)=q_i$/

claim: h satisfies the required conditions for any n.

let n=1, then $h(p_1)=q_1$. suppose h satisfies the condition for n=k, i.e., $h(p_i)=q_i$ for values of i in $\{0,1,...,k\}$.Then for $n=k+1$, $h(p_i)=q_i$ for values of $i \in \{0,1,...,k,k+1\}$ so h satisfies the conditions for any n.

construct concatenations using the natural maps

$f:X/(p_i\sim p_i)\rightarrow [0,1]$
$g:Y/(q_i\sim q_i)\rightarrow [0,1]$

h is continuous because for all open sets in $Y/(q_i\sim q_i)$, the inverse image is an open set in $X/(p_i\sim p_i)$.
the inverse of h is continuous because for all open sets in $X/(p_i\sim p_i)$, there is an open set in $Y/(q_i\sim q_i)$.
h is onto and one-to one because $Y/(q_i\sim q_i)$ is covered once by the image of $X/(p_i\sim p_i)$.

so h is a homeomorphism.