<s>[INST] Proving that $(a,b)\cong [1,2)$. I aim to prove the proposition indicated in the title. 
Is the following argument correct?
We now show that the map $h:(a,b)\to [1,2)$ defined as follows is a bijection.
\begin{align*}h(x) = \begin{cases}\frac{x-a}{b-a}+1&\text{ if }x\neq \frac{a+b}{2}\\1&\text{ if }x = \frac{a+b}{2}\\\end{cases}\end{align*}
Let $x\in(a,b)$. Assume for the purpose of contradiction that $h(x) = 1$ and $x\neq\frac{a+b}{2}$. Now $x>a$ and $b>a$, implying $\frac{x-a}{b-a}>0$, so $h(x) = \frac{x-a}{b-a}+1>1$. In summary $h(x)=1$ if and only if $x=\frac{a+b}{2}$.
Now let $x_1,x_2\in(a,b)$ and assume $h(x_1) = h(x_2)$ now if $h(x_1) = h(x_2)$ then $x_1 = x_2 = \frac{a+b}{2}$ and if $h(x_1) = h(x_2)\neq 1$, then both $x_1$ and $x_2$ are not equal to  $\frac{a+b}{2}$, thus $h(x_1) = \frac{x_1-a}{b-a}+1 = \frac{x_2-a}{b-a}+1= h(x_2)$, implying $x_1 = x_2$.
Let $y\in[1,2)$. In the event $y = 1$, then $h(\frac{a+b}{2}) = 1$ and if $y\neq 1$, then given the injectivity of $h$, we may have $h(x) \neq 1$ only if $x\neq\frac{a+b}{2}$. Let $x = y(b-a)+a$, here $x\neq \frac{a+b}{2}$ as arguing to the contrary would imply that $y = \frac{1}{(b-a)}(\frac{a+b}{2}-a) = \frac{1}{2}$, which is impossible considering $y\in[1,2)$. Then given the definition of $h$ it is then apparent that $h(x) = y$. Thus the map in question is surjective.
$\blacksquare$
 [/INST] First, note that your function is not surjective, since there is no value in $(a,b)$ that maps to $\frac32$. We know this because $\frac{x-a}{b-a}+1$ is continuous and strictly increasing on the set $(a,b)$. As such, we know that there is only one value on the set that can map to $\frac32$, which is $x=\frac{a+b}2$. However, $\frac{a+b}2$ maps to $1$.
Instead, we note that since the composition of bijective maps is bijective, we can split the problem into finding bijective maps of the following kind:
$$f: (a,b)\to(0,1)$$$$g: (0,1)\to[0,1)$$$$h:[0,1)\to[1,2)$$
Here are the mappings we will use $$f(x)=\frac{x-a}{b-a}$$$$g(x)=\begin{cases}\frac{n-1}n&\exists n\in\mathbb N,\;x=\frac n{n+1}\\x&\textrm{otherwise}\end{cases}$$$$h(x)=x+1$$
It's obvious that the composition of these functions gives you your desired function. Can you prove these functions are bijective?
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