## Tag Archives: injective

### Injections, surjections and bijections

A function $f: A \rightarrow B$ is called an injection if $f(a)=f(a^{'})$ implies $a=a^{'}$. There is another name for this kind of function. It is also called a one-to-one function or an injective map. A map $f:A \rightarrow B$ is one-one if two distinct elements of A have distinct images under f.

A function $f: A \rightarrow B$ is called a surjection if for every $b \in B$ there is an element $a \in A$ such that $f(a)=b$. In other words, $f(A)=B$. That is to say, every element of B is an image of some element of A under f. A surjective map is also called an onto map.

A map $f:A \rightarrow B$ which is both one-to-one and onto is called a bijection or a bijective map.

Examples.

1) Suppose $f: \Re \rightarrow \Re$ is defined by $f(x) =\cos{x}$. It is clear that this is neither one-to-one nor onto. Indeed because $f(x)=f(x+2\pi)$, it cannot be one-to-one. Since $f(x)$ never takes a value below -1 or above 1, it cannot be onto.

2) $f: \Re \rightarrow \Re$ defined by

$f(x)= x$, if $x \geq 0$ and $f(x)=x-1$, if $x<0$ is one-to-one but not onto as $-1/2$ is never attained by the function.

3) $f:\Re \rightarrow \Re$ defined by $f(x)=\frac{x}{1+|x|}$ is one-to-one but not onto.

Warning.

Trigonometric functions like sine and cosine are neither one-to-one nor onto. So, how does one define $\sin^{-1}{x}$ or $\cos^{-1}{x}$? Actually, there is ambiguity in defining these. If we write $\sin^{-1}{x}=\theta$, it means that $\sin{\theta}=x$. It is easily seen that there is no $\theta$ if x is more than 1 or less than -1. Thus, the domain of $\sin^{-1}{x}$ or $\cos^{-1}{x}$ must be $[-1.1]$. Then, again

$\sin{\theta}=x$ has many solutions $\theta$ for the same x. For example, $\sin{\pi/6}=\sin{5\pi/6}=1/2$. So, which of $\pi/6$ or $5\pi/6$ should claim to be the value of $\sin^{-1}{(1/2)}$? In such a case, we agree to take only one value in a definite way. For $0 \leq x \leq 1$, we choose $0 \leq \theta \leq \pi/2$ such that $\sin{\theta}=x$. It is obvious that there is only one such $\theta$. Similarly, for $-1 \leq x <0$, we choose $-\pi/2 \leq \theta < 0$ such that $\sin{\theta}=x$. Thus, this way of choosing $\theta$ such that $\sin{\theta}=x$ for $-1 \leq x <1$ has no ambiguity. such a value of the inverse circular function is called its principal value, though we could choose another set of values with equal ease.

Similar problems arise in the context of a function $f:\Re \rightarrow \Re$ defined by $f(x)=x^{2}$. The function f is neither one-to-one nor onto. But, if we take $f: \Re \rightarrow \Re_{+}$ as $f(x)=x^{2}$, then f is onto. We would like to define $f^{-1}:\Re_{+} \rightarrow \Re$ as a function. In order that we are able to define $f^{-1}$ as a function we must agree, once and for all, the sign of $f^{-1}{(x)}$. Indeed, since $f(-1)=f(1)=1$, which one would we call $f^{-1}$ ? In fact, $f^{-1}{(x)}$ is what we would like to denote by $\sqrt{x}$. But, we must decide if we are taking the positive value or the negative value. Once, we decide that, $f^{-1}$ would become a function.

More later,

Nalin Pithwa