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?
On
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.
On
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 $f(x) = y$.
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 $Y$.