If a function has an inverse then it is bijective? I have some trouble finding the answer to this, can someone help me out:
If I have a general function $f$ with domain $X$ and codomain $Y$, I know nothing about the function (injective, surjective). Say I have found and inverse that is, there is a function $g$ such that for $f(x) = y$, $g(y) = x$ (is this enough for an inverse). Then the function is bijective?
 A: Not necessarily. Simply, the fact that it has an inverse does not imply that it is surjective, only that it is injective in its domain. 
Take for example the functions $f(x)=1/x^n$ where $n$ is any real number. As $x$ approaches infinity, $f(x)$ will approach $0$, however, it never reaches $0$, therefore, though the function is inyective, and has an inverse, it is not surjective, and therefore not bijective.
A: If you have $f: X \to Y$, $g : Y \to X$ with $g(f(x)) = x$ for all $x \in X$, it is still possible for $f$ to not be bijective.
However, $f$ will be a bijection onto its image; i.e., $f$ is a bijection from $X$ to $f(X)$. In other words, $f$ is injective.
If you additionally require that $f(g(y)) = y$ for all $y \in Y$ (i.e. $g$ is a "two-sided inverse"), then $f$ is a bijection from $X$ to $Y$.
A: This is not true in general. The fact that $f$ has an inverse means the function is injective but for it to be bijective it needs to be surjective as well.
Let's see a simple example where $f$ is discrete.
Let the domain and codomain be $X:=\{1,2\}$ and $Y:=\{2,4,6\}$. Define the function $f: X\mapsto Y$ as follows
$$f(1) = 2, f(2)=4$$
The inverse function is simple: $$f^{-1}(2) = 1, f^{-1}(4)=2$$
but $f$ is not a bijection because $X$ and $Y$ do not have equal cardinality.
Remember that the function $f: X \mapsto Y$ is bijective iff for all $y \in Y$, there is a unique $x \in X$ such that $f(x) = y$.
A: If you have f:X→Y,f:X→Y, g:Y→X ,g:Y→X with g(f(x))=x   g(f(x))=x for all x∈X  x∈X, it is still possible for ff to not be bijective. However, ff will be a bijection onto its image; i.e., ff is a bijection from XX to f(X)f(X). In particular, ff is injective.
If you additionally require that f(g(y))=yf(g(y))=y for all y∈Yy∈Y (i.e. gg is a "two-sided inverse"), then ff is a bijection from XX to YY.
