I'm working out of the Nakahara text in mathematical physics, and I'm presented with a map $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $ f:x \mapsto \sin(x) $, and told that it is neither injective nor surjective, and to restrict the domain and range to make the mapping bijective. Redfining the map in more general terms:
$$ X,Y \subseteq \mathbb{R} \,\, | \,\, f: X \rightarrow Y $$
I first sought to investigate how this mapping fails to be injective. Knowing the definition of injectivity, I immediately see that there exists more than one possible inverse value for every value of the range. In other words, I can draw a line of constant $y$ on the graph of this function and strike the function more than one time, and in fact infinitely many times as $x$ is allowed to run to infinity. This leads me to believe I need to put a restriction involving the periodicity of $\sin(x)$ on the domain for sure.
Secondly, I checked how the mapping fails to be surjective. I know the domain runs from negative infinity to infinity, whereas the range only runs from $-1$ to $1$. Surjectivity requires that for every $ y \in Y $, there exists at least one $ x \in X$ such that $ f(x) = y$. If $ Y$ is the entire real line and the range of $f$ is $[-1,1]$, then any $y$ outside of that interval will fail to have an inverse image, which contradicts surjectivity.(?)
To achieve injectivity, I considered the fact that $sin(x)$ is periodic, with a first maximum at $\frac{\pi}{2}$. Immediately after that, we start seeing repeated $y$ values for different $x$ values. I also noticed that we can extend this towards negative $x$ out to the same value. For surjectivity, and thus bijectivity, I noted that we need every value in both the domain and range to be utilized in the mapping. We explicitly have the required range.
Using the aforementioned reasoning, I conclude that the mapping with restrictions should look like this: $$ X = [-\frac{\pi}{2},\frac{\pi}{2}] \,\, , \,\, Y = [-1,1]$$ $$ f: X \rightarrow Y \,\, | \,\, f: x \mapsto \sin(x) $$
Is all that rationale provided above sufficient to make this map bijective?