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Inverse Limit Space Homeomorphism

joypav

Active member
Mar 21, 2017
151
Problem:
Suppose that $X = lim_{\leftarrow}\left\{X_i, f_i\right\}_{i=1}^{\infty}$ is an inverse limit space so that there is an integer $N$ and for each $n ≥ N$ the function $f_n$ is an onto homeomorphism. Then $X$ is homeomorphic to $X_N$.

Solution:
It seems simple that the map we are looking for is
$f: X \rightarrow X_N$
for $P=\left\{P_i\right\}_{i=1}^{\infty} \in X, f(P)=P_N$
(However I may be wrong and it wasn't so simple...)

Its inverse.
I have defined the inverse of $f$ as follows...
Let $f^{-1}=g$. (Note: $g(x)_k$ means the $k$th coordinate of $g$)
For $x \in X_N$,

.
.
.
$g(x)_{N+k}=f_{N+k-1}^{-1} \circ ... \circ f_N^{-1}(x)$
.
.
.
$g(x)_{N+2}=f_{N+1}^{-1}(f_N^{-1}(x))$
$g(x)_{N+1}=f_N^{-1}(x)$
$g(x)_N=x$
$g(x)_{N-1}=f_{N-1}(x)$
$g(x)_{N-2}=f_{N-1}(f_{N-1}(x))$
.
.
.
$g(x)_{N-k}=f_{N-k} \circ ... \circ f_{N-1}(x)$
.
.
.

Claim: $g$ and $f$ are inverses

1. Show: $f(g(P_N))=P_N$
Consider $P_N \in X_N$.
$g(P_N)=(..., g(P_N)_{N-1}, g(P_N)_N, g(P_N)_{N+1},...) = (..., f_{N-1}(P_N), P_N, f_N^{-1}(P_N),...) = \left\{P_i\right\}_{i=1}^{\infty}$
$\implies f(g(P_N)) = f(\left\{P_i\right\}_{i=1}^{\infty}) = P_N$

I am leaving out some details here. But $g$ should map this point into a point of $X$, satisfying the necessary properties.
Those being that $P = \left\{P_i\right\}_{i=1}^{\infty}$ is a point of $X$ provided for each positive integer $i, P_i \in X_i$ and $f_i(P_{i+1}) = P_i$.

2. Show: $g(f(P))=P$
Consider $P \in X$.
$f(P) = P_N$.
Then $g(f(P))=g(P_N)=P$.

1. and 2. $\implies f$ and $g$ are inverses.

Claim: $g$ is continuous

$g$ is composed of continuous functions (every $f_i$ is continuous by definition of the inverse limit space) and therefore $g$ is continuous.

Claim: $f$ is continuous
I am stuck on this one. I need to show that $g$ is an open map. Meaning, for $U_N$ open in $X_N, g(U_N)$ is open in $X$.
Suggestions??

Claim: $f$ is onto
To Show: $\forall P_N \in X_N, \exists P \in X$ such that $f(P)=P_N$
Choose $P=g(P_N)$.

Claim: $f$ is one-to-one
Again, I am stuck on this one. I can only get so far.
Suppose we have $P^i, P^j \in X$.
To Show: $f(P^i)=f(P^j) \implies P^i=P^j$
Suppose $f(P^i)=f(P^j)$
$\implies P_N^i=P_N^j$
$\implies f_N(P_{N+1}^i)=f_N(P_{N+1}^j)$ (because the definition of a point in $X$ is that each component must satisfy $f_i(P_{i+1})=P_i$)
$\implies P_{N+1}^i=P_{N+1}^j$ (because for every $n \geq N, f_n$ is a homeomorphism and is therefore one to one)
$\implies f_{N+1}(P_{N+2}^i)=f_{N+1}(P_{N+2}^j)$
$\implies P_{N+2}^i=P_{N+2}^j$
.
.
.
Continue in this way.
This shows that $P^i=P^j$ from the $N$th coordinate on. What about before the $N$th coordinate? Can we ensure that they will be equal?
Does the existence of our inverse $g$ imply bijectivity?
 
Last edited:

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
Hi joypav ,

For clarification, what do you mean when you say "the function $f_n$ is an onto homeomorphism?" The standard definition of homeomorphism already implies surjectivity.
 

joypav

Active member
Mar 21, 2017
151
Hi joypav ,

For clarification, what do you mean when you say "the function $f_n$ is an onto homeomorphism?" The standard definition of homeomorphism already implies surjectivity.
I think it may have just been for emphasis? I'm not sure. I just stated it as it was in our notes.. but yes it's redundant!

Here is what I ended up doing. I'm thinking it does work out okay..

The inverse is defined as I stated. I think my reasoning for $g$ being continuous and $f$ onto is okay.

Claim: $f$ is continuous
To Show: $U_N$ open in $X_N \implies f^{-1}(U_N)=g(U_N)$ is open in $X$
Claim: $g(U_N)=U_N^{\leftarrow}$, where $U_N^{\leftarrow}=\{P \in X: P_N \in U_N\}$
(this would then show that $g(U_N)$ is open in $X$, because $U_N^{\leftarrow}$ is a basis element for the topology on $X$)

1. $g(U_N) \subset U_N^{\leftarrow}$
Take $P \in g(U_N)$
$\implies f(P) \in U_N$
$\implies P_N \in U_N$
$\implies P \in U_N^{\leftarrow}$

2. $U_N^{\leftarrow} \subset g(U_N)$
Take $P \in U_N^{\leftarrow}$
$\implies P_N \in U_N$
$\implies g(P_N) \in g(U_N)$
$\implies P \in g(U_N)$

1. and 2. $\implies$ Claim $\implies f^{-1}(U_N)=g(U_N)$ is open in $X \implies$ Claim

Claim: $f$ is one to one

Recall that...
"The element $P = \{P_i\}_{i=1}^{\infty}$ is a point of $X$ provided for each positive integer $i, P_i \in Xi$ and $f_i(P_{i+1}) = P_i$."

To Show: $f(P^j)=f(P^i) \implies P^j=P^i$

$f(P^j)=f(P^i) \implies P_N^j=P_N^i \implies$ (because of what is in quotations) $f_N(P_{N+1}^j)=f_N(P_{N+1}^i) \implies P_{N+1}^j=P_{N+1}^i$
... continue in this manner.
Then $P_m^j = P_m^i$ for all $m \geq N$.

Using our condition for the $f_i$'s in quotations again, we have,
$f_{N-1}(P_N^j)=P_{N-1}^j$ and $f_{N-1}(P_N^i)=P_{N-1}^i \implies P_{N-1}^j=P_{N-1}^i$
Then, $P_m^j = P_m^i$ for all $m < N$.

So we can conclude that $P^j=P^i$, (because every coordinate is equal).