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?


2 Answers 2


Yes, you have provided both the correct rationale and almost (see my last paragraph) provided valid restrictions to make the function bijective. It is worth noting that these restrictions aren't unique, for example we could instead take $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$ for injectivity.

Regarding this, you may find this to be of interest.

I would like to make a point regarding your notation - $X$ is the domain, so the intention is to choose it explicitly. In your answer, it should be $X = [-\frac{\pi}{2}, \frac{\pi}{2}]$, rather than $X \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, otherwise what we're saying is that the entire domain $X$ is a single element of $[-\frac{\pi}{2}, \frac{\pi}{2}]$. And similarly for your $Y$.

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    $\begingroup$ Edits noted and internalized! $\endgroup$ Apr 14, 2016 at 7:49
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    $\begingroup$ The function $f(x) = \sin x$ is not injective on the domain $[0, \pi]$. Note that $\sin 0 = \sin \pi = 0$. More generally, $\sin x = \sin(\pi - x)$, so every $y$-value in the interval $[0, 1]$ appears twice except $1 = \sin(\frac{\pi}{2})$. $\endgroup$ Apr 14, 2016 at 9:51
  • $\begingroup$ @N.F.Taussig You are correct, I have edited my post to reflect this (I was thinking of $\cos$ rather than $\sin$!) $\endgroup$ Apr 14, 2016 at 9:52

Yes, your reasoning is correct, although the answer is not the only possible.

Defining the function as $$(0,\tfrac\pi 2] \to (0,1]$$ or $$(\tfrac\pi 2,\pi] \to [0,1)$$ or even $$\{0\} \to \{0\}$$ satisfies requirements, too.


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