Given the CDF
$$F_X(x)= \begin{cases} \frac12 e^x&,\text{ if }x<0 \\ \frac12&,\text{ if }0\le x<1 \\ 1-\frac12 e^{1-x}&,\text{ if }x\ge 1 \end{cases}$$
I am trying to find $F_X^{-1}(y)$ of this CDF.
For $e^x/2$, I got $\ln(2y)$ and the interval becomes $x<0 \implies \ln2y<0\implies y<1/2$.
For $1-(e^{1-x}/2)$, I got $1-\ln(2(1-y))$ and the interval becomes $1/2 \leq y$
My question is, the solution gives intervals $[0,1/2)$ for the first one and $[1/2,1)$ for the second one. Where did we get these bounds? Do they come from the definition of $\ln$? If so should we not include 0 because $\ln0$ doesn't makes sense? Also, at which step we dealt with 1/2 in the CDF? How it effects the inverse if it effects at all? Thanks a lot in advance!
The "inverse" of the CDF is called the quantile function. It is really a generalized inverse because it exists even when $F$ is not a bijection from $[-\infty,\infty]$ to $[0,1]$. In this situation it is not an inverse because your CDF is not one-to-one.
That said, you have $F_X^{-1}(y)$ correct for $0<y<1/2$ and $1/2<y<1$. For $y=0$ and $y=1$ really the only sensible way to define $F_X^{-1}(y)$ is as $-\infty$ and $+\infty$ respectively, so that monotonicity is retained.
For $y=1/2$ the definition of $F_X^{-1}(y)$ becomes sensitive to how exactly you choose to define the quantile function. The most common definition is $F_X^{-1}(y)=\inf \{ x : y \leq F_X(x) \}$, so that here $F_X^{-1}(1/2)=1$.