My Professor instructed us to read a textbook which opens with the following explanation:
Another way of looking at the Laplace transform is as a mapping from points in the $t$ domain to points in the $s$ domain. Pictorially, Fig. 1.1 indicates this mapping process. The time domain $t$ will contain all those functions $F(t)$ whose Laplace transform exists, whereas the frequency domain $s$ contains all the images $\mathcal{L}\{F(t)\}$.
Dyke, Phil. An Introduction to Laplace Transforms and Fourier Series (Springer Undergraduate Mathematics Series)
What I find confusing is that the author says that the time domain $t$ will contain all those function $F(t)$. But $F(t)$ is not in $t$: $t$ is the domain of $F(t)$ (which, as I understand it, is $(0, \infty)$), and the set of all values $F(t)$ would be separate from that?
And then, if $T$ is the domain and $Y$ is the codomain, then the set of all images are $\{y \in Y \mid y = F(t) \ \forall \ t \in T \}$. So how can the author say that the frequency domain $s$ contains all the "images" $\mathcal{L}\{F(t)\}$?
Something about this explanation doesn't seem correct.
Best Answer
"$t$ domain" or "time domain $t$" in your textbook, does not mean the domain of the specific function $F(t)$. Rather, it means a set $T := \{F(t)|F(t)\textrm{ is a function which is defined on time, namely on }(0,\infty)\}$. Such set of function has somewhat space-like structure, so $T$ is also called a space. By the same way, "frequency domain $s$" means $S := \{G(s)|G(s)\textrm{ is a function which is defined on frequency, namely on }(0,\infty)\}$.
So, with this notations, $\mathcal{L}\{T\} \subset S$. That's what author meant.
In reallity, the definition I used above is not that exact: it needs further formality. But I am sure you get familiar with it as you study further.