To me this seems tautological. But I've had someone challenge me on it. Here $\mathbb{R}$ is the set of real numbers.
What I call the domain of a mapping is the set of all arguments for which there is a value of the mapping. For example, consider the mapping $f:\mathcal{D}\to{R}$ whith $f\left(x\right)\equiv\sqrt{x},$ and $x,f\left(x\right)\in{\mathbb{R}}$. The domain of $f$ is $\mathcal{D}=\{x\in\mathbb{R}\backepsilon x\ge 0\}$. The image (range) is $\mathcal{R}=f\left(\mathcal{D}\right)=\mathcal{D}.$ That is, the domain and image are identical. Since $f$ is monotonic, it is one-to-one. Thereofore $f:\mathcal{D}\to{R}$ is a bijection. The image and co-domain are identical.
On the other hand if $g\left(x\right)\equiv\sqrt{x},$ the mapping $g:\mathcal{D}\to{\mathbb{R}}$ is an injection, but not a bijection. So the determination as to whether the mapping is a bijection depends on the specification of the co-domain.
The domain of the mapping is the same in both cases, and it is incorrect (but not too uncommon) to write $g:\mathbb{R}\to{\mathbb{R}}$ or $f:\mathbb{R}\to{\mathbb{R}}.$ The image of $g$ is still $\mathcal{R}.$
Is this correct?
This is "mostly correct" - there are two points to be made.
The first concerns function identity and your use of the word "is." This is actually an important point.
In many (most?) presentations, a function is determined by three pieces of data: its domain $D$, its codomain $C$, and the corresponding "graph" relation $\subseteq D\times C$ (namely, $\{(a,b): f(a)=b\}$). In this context, changing the codomain changes the function: e.g. $f: \mathbb{R}\rightarrow\mathbb{R}: x\mapsto x^2$ is a different function from $g: \mathbb{R}\rightarrow \mathbb{R}_{\ge 0}: x\mapsto x^2$. And so the statement "every one-to-one mapping is a bijection between its domain and its image" would be technically incorrect. The right formulation of the obvious fact would be a bit more complicated - at the very least, one probably ought to say "yields" (or "induces," or etc.) instead of "is."
In other contexts - e.g. set theory - a function is determined entirely by (indeed, is) its graph $\{(a,b): f(a)=b\}$, and according to this approach your statement as phrased is correct.
The other issue concerns terminology - specifically, a terminological issue which arises if you phrase your question slightly differently (and if you happened to phrase it differently in your conversation you mentioned, it could actually be relevant). This isn't really important.
The word range is (to my understanding) used to refer to the image in most modern texts, but some texts (such as Munkres) use it to refer to the codomain. So if you had said "every one-to-one mapping is a bijection between its domain and its range," your interlocutor might object if they interpret "range" as "codomain."