Topological isomorphism, inverse function

Question

$c:=\{f:\mathbb{N}\to\mathbb{C}\colon \lim_{n\to\infty} f(n)~ \text{exists}\}$ and $c_0:=\{f:\mathbb{N}\to\mathbb{C}\colon \lim_{n\to\infty} f(n)=0\}$ both with $||f(n)||_\infty=\sup_{n\in\mathbb{N}} |f(n)|$. Let $T: c_0\to c$ with $(Tf)(n):=f(n+1)+f(1)$ for all $n\in\mathbb{N}$.

Show, that $T$ is a topological isomorphism.

To show that $T$ is a topological isomorphism first of all $T$ has to be linear. Then $T$ must be bijective and $T$ as well as $T^{-1}$ have to be continuous.

That $T$ is linear is clear. For $f,g\in c$ and $\mu, \lambda\in\mathbb{K}$ it is:

$(T(\lambda f+\mu g))(n)=(\lambda f+\mu g)(n+1)+(\lambda f+\mu g)(1)\\=(\lambda f)(n+1)+(\lambda f)(1)+(\mu g)(n+1)+(\mu g)(1)\\=(T(\lambda f))(n)+(T(\mu g))(1)$

Now I want to find $T^{-1}$. Do I have to find $T^{-1}$ such that:

$T^{-1}((Tf)(n))\stackrel{?}{=}f(n)$ and $T((T^{-1}f)(n))\stackrel{?}{=}f(n)$

But I do not succeed... Can you give a hint? Thanks in advance.

Answer 1

Hint: Fix $f\in c_0$. Then $$\lim_{n\to\infty}Tf(n)=f(1)+\lim_{n\to\infty}f(n+1)=f(1)$$ Now suppose $g\in c$. What should $T^{-1}g(1)$ be?

Answer 2

The map $T$ can be seen as $$ T(a_1,a_2,\ldots)=(a_2+a_1,a_3+a_1,a_4+a_1,\ldots) $$ To "come back", you need to put the limit of the sequence in the first coordinate, and subtract it from the other entries: if we write $b_\infty$ for the limit of the sequence $(b_n)$, $$ T^{-1}(b_1,b_2,\ldots)=(b_\infty, b_1-b_\infty,b_2-b_\infty,\ldots) $$