Hello. Does it always an inverse map don't exist? I know that a mapping is considered as a function . And I know that there are some functions as sine that has an inverse that is arcsin . So why might some functions won't have an inverse or am I wrong? Another thing is why shouldn't the notation of the inverse map be broken into component parts?
Note: the pic is taken from the book " A course in modern mathematical physics" for peter scekerz. Chapter one , section 1.4 Mappings .
On
Let $A= \{0,1\}$ and $B=\{0\}$. Consider the function $f: A \to B$ defined by $f(0)=f(1)=0$. This function has no inverse because the inverse would have to send the element $0 \in B$ to both $0$ and $1$ in $A$, and this is not a function.
In general, a function is has an inverse if and only if it is injective and surjective. In the example I gave, $f$ is surjective, but not injective.
A function $f$ from a set $X$ (called domain) into a set $Y$ (called codomain) is a rule that assigns every $x$ in $X$ to a unique $y$ in $Y$, where $y$ is usually denoted by $f(x)$. The function $f$ has an inverse if and only if it makes sense to reverse the rule on $f$. That is, if and only if for every $y$ in $Y$ there is a unique $x$ in $X$ such that $f(x)=y$. With this, there are two ways that the function can fail to have an inverse.
1) The function $f$ may map multiple elements of the domain to the same point in the codomain. As an example, suppose $f:\{0,1\}\to\{a\}$ is defined by $f(0)=f(1)=a$. This function doesn't have an inverse because defining $g:\{a\}\to\{0,1\}$ by $g(a)=0$ and $g(a)=1$ is not a valid function. Indeed, this $g$ does not satisfy the uniqueness part of the definition (or, as you may be familiar with, the "vertical line test"). This doesn't work precisely because $f$ is not injective.
2) The function $f$ may not map to all elements of the codomain, so there is no way to define an inverse on the entire codomain. As an example, suppose $f:\{0\}\to\{a,b\}$ is defined by $f(0)=a$. This function does not have an inverse because defining $g:\{a,b\}\to\{0\}$ by $g(a)=0$ is not a valid function. To see this, note that $b$ is an element of the domain of $g$ that is left unassigned. This doesn't work precisely because $f$ is not surjective.
In fact, these are exactly the only two reasons that $f$ can fail to have an inverse. One can prove that $f$ has an inverse if and only if it is both injective and surjective.