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**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 $x \in X_N$,

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

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$g(x)_{N+2}=f_{N+1}^{-1}(f_N^{-1}(x))$

$g(x)_{N+1}=f_N^{-1}(x)$

$g(x)_N=x$

$g(x)_{N-1}=f_{N-1}(x)$

$g(x)_{N-2}=f_{N-1}(f_{N-1}(x))$

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$g(x)_{N-k}=f_{N-k} \circ ... \circ f_{N-1}(x)$

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**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)

$\implies f_{N+1}(P_{N+2}^i)=f_{N+1}(P_{N+2}^j)$

$\implies P_{N+2}^i=P_{N+2}^j$

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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?

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