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Here is my previous question Investigating the relationship between $X$ and the inverse limit $\mathcal{L}$ of the tower.

And now I want to Find a way to uniquely determine a point $x_{\sigma} \in X,$ given an element $\sigma \in \mathcal{L}.$ This defines a function $f:\mathcal{L} \rightarrow X.$

Could anyone help me in doing so please?

Idonotknow
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1 Answers1

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As I said in my answer, we can define it by $\{x_\sigma\} = \bigcap_n \sigma_n$.

The map is 1-1 because if $\sigma \neq \sigma'$, then for some $n$ we have $\sigma_n \neq \sigma'_n$ and as these are from the same $\mathcal{A}_n$, which is a partition, $\sigma_n \cap \sigma'_n = \emptyset$ which implies $x_\sigma \neq x_{\sigma'}$.

The map is also onto, as each $x$ occurs in a unique $\sigma_n \in \mathcal{A}_n$ and then the $\sigma$, so defined, has $x_\sigma=x$ by definition.

The last function also gives the inverse $g: X \to \mathcal{L}$: $g(x)=\sigma$ defined by: $\sigma_n$ is the unique member of $\mathcal{A}_n$ that contains $x$.

Henno Brandsma
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  • And how can I show that the 2 functions are inverses of each other? –  Apr 11 '20 at 22:31
  • I know that I should show that $f \circ g = id_{X}$ and $g \circ f = id_{\mathcal{L}}$ but how? –  Apr 11 '20 at 22:36