On a bijective Laplace transform

[psie]

TL;DR

I've been going in circles with my reasoning about this, so I hope this doesn't come across as too circular. I'm reading Ordinary Differential Equations by Adkins and Davidson and in the chapter on Laplace transforms, there are two sections where they establish a one-to-one correspondence of the Laplace transform between two function spaces. Everything leads up to a theorem in which they prove the one-to-one correspondence, but I don't see the need for the proof.

Theorem. The Laplace transform $\mathcal{L}:X\to Y$ establishes a linear one-to-one correspondence between the linear space of exponential polynomials, $X$, and the linear space of proper rational functions, $Y$.

Exponential polynomials are linear combinations of terms like $t^ne^{at}\cos{bt}$ and $t^ne^{at}\sin{bt}$, where $n$ is a nonnegative integer, $a$ is real and $b>0$. Like the proper rational functions, these are presumably subspaces of $\mathbb R^{\mathbb R}$.

The proof goes like this (I have omitted it since it refers to quite a few previously established results, but the point is the logic of the proof). The authors show that $\mathcal{L}$ maps each exponential polynomial to a proper rational function and that a function, denoted by $\mathcal{L}^{-1}$, and defined earlier by $$\mathcal{L}^{-1}(F(s))=f(t)\iff \mathcal{L}(f(t))=F(s),\tag1$$ maps each proper rational function to an exponential polynomial. Moreover, they note that $\mathcal{L}$ is injective for continuous functions, which includes exponential polynomials. Then they conclude that $\mathcal{L}$ between these two spaces is bijective.

Here is what puzzles me. If a function has an inverse, as defined by $(1)$, it's automatically a bijection. So it seems to me there is nothing to prove. $(1)$ applies to any function that has an inverse, with a suitable change of letters, inputs and outputs. Nowhere in the chapter so far have they explicitly defined $\mathcal{L}^{-1}$ and shown $\mathcal{L}^{-1}\circ \mathcal{L}=\mathrm{id}_X$ and $\mathcal{L}\circ \mathcal{L}^{-1}=\mathrm{id}_Y$, where $\mathrm{id}_A$ denotes the identity map on some set $A$. These observations simply puzzle me and I'd be grateful if someone could share their viewpoint on the matter.

 

[pasmith]

By introducing \mathcal{L}^{-1} through <br /> \mathcal{L}^{-1}(F(s))=f(t)\iff \mathcal{L}(f(t))=F(s)<br /> the authors are assuming, so far withour justification, that \mathcal{L} is actually invertible. The theorem provides that justification, at least for particular types of functions.