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# Bijections and compositions

## Bijection

A function $f:X\to Y$ is bijective if, for each value $y\in Y$, there is exactly one element $x\in X$ that is mapped to $y$. In other words, the equation $$y=f(x)$$ has a unique solution $x$ for every $y$.

We call $x$ the pre-image of $y$ and we denote it $f^{-1}(y)$. We usually compute the pre-image of $y$ by expressing $x$ in terms of $y$.

A bijective function is also called a bijection or a one-to-one correspondence.

The function $x^2$ defined on $\R$ is not bijective, because $1$ has two pre-images ($f(-1) = f(1) = 1$).

The function $x^3$ is bijective on $\R$, because every $y$ in $\R$ has a unique pre-image, specifically $x=y^{1/3}$.

I and II are functions, but III is not. I is a bijection. II is not a bijection.

## Composition of functions

The composition of functions is the successive application of functions. The resulting function is called a composite function.

The composite of $f(x)=2x$ and $g(x)=x^2$ is $h(x)=f(g(x))=2x^2$.

More formally, if $g:D_g\to R_g$ and $f:D_f\to R_f$ are two functions with $$R_g\subset D_f,$$ the composite function $f\circ g$ is the function from $D_g$ to $R_f$ defined by $$f\circ g (x) = f\big(g(x)\big).$$

The composite function is written either as $f\circ g$ or $fg$ and is read $f$ of $g$. $fg$ is often used for the product of functions, so $f\circ g$ should be preferred.

$h(x) = \sin(x^2)$ is the composite $\sin\circ g$ where $g(x) = x^2$. $\vert x^3\vert$ is the composite function $f\circ g$ with $f(x) =\vert x\vert$ and $g(x) = x^3$.

## Inverse function

The inverse function of a bijection $f:X\to Y$ is the function $f^{-1}:Y\to X$ that maps each value to its unique pre-image $$y=f(x) \Longleftrightarrow x=f^{-1}(y).$$

The inverse of $x^2$ from $[0, +\infty)$ to $[0, +\infty)$ is $\sqrt{x}$. This is because $y=x^2$ is equivalent to $x = \sqrt{y}$. The inverse of $\ln x$ is $e^x$.

The domain of the inverse $f^{-1}$ is the range of the original function $f$. The range of the inverse $f^{-1}$ is the domain of $f$. To summarise, $$D_{f^{-1}} = R_f,\quad R_{f^{-1}} = D_f.$$

$(0,+\infty)$ is the range of $e^x$ and the domain of $\ln x$.

The inverse function is bijective and its inverse is the original function $$(f^{-1})^{-1} = f.$$

The graph of the inverse function is deduced from the graph of the function by a reflection across the line $y=x$. This is because the two graphs simply have the $x$ and $y$ axes inverted.

Graph of $f^{-1}$ and graph of $f$

## Composition is not commutative

Contrary to the properties of sums and products, order matters in composition. In general, $$f\circ g\ne g\circ f.$$ We say that composition is not commutative.

Setting $f(x) = x^2$ and $g(x) = x+1$, we see that, for $x\ne0$, $$f\circ g(x) = (x+1)^2\ne x^2+1 = g\circ f(x).$$

However, for the inverse function, we have $$f\circ f^{-1}(y) =y,\qquad f^{-1}\circ f(x) =x$$ This comes from $y=f(x)\Longleftrightarrow x=f^{-1}(y)$.

The composition of monotonic functions is monotonic. If both functions have same monotonicity (both increasing or decreasing), the composite function is increasing; otherwise it is decreasing.

To see this, assume that both $f$ and $g$ are decreasing. Then, for $x\le y$, we have $g(x)\ge g(y)$, hence $f\left(g\left(x\right)\right)\le f\left(g\left(x\right)\right)$. This means that means that $f\circ g$ is increasing.