Is $f(x) = e^x$ one-to-one if $f:\mathbb{R} \rightarrow \mathbb{R}$? My book says that $f(x) = e^x$ is not invertible from the set of real numbers to the set of real numbers. But I disagree since $f(x) = e^x$ is injective with this given domain and codomain and therefore invertible.
 A: $f(x)=e^x$, with $f:\mathbb R \to \mathbb R$, is injective but not surjective since $e^x>0$ for all $x \in \mathbb R$.
To turn an injective function into a bijective (hence invertible) function, it suffices to replace its codomain by its actual range. In this particular case that means defining the function as $f:\mathbb R \to \mathbb R_{++}.$
A: $f: \mathbb R\to (0,\infty)$, defined as $f(x)=e^x$ is bijective. there exists an inverse function $f^{-1}:(0,\infty)\to\mathbb R$ for which we have $$f^{-1}(f(x)) = x$$ for all $x\in\mathbb R$ and $f(f^{-1}(x))=x$ for all $x\in(0,\infty)
However, $g:\mathbb R\to\mathbb R$, defined as $g(x)=e^x$ is not bijective, because there is no $x\in\mathbb R$ such that $g(x)=-1$.
A: Perhaps the author has in mind that $\exp : \mathbb R \to (0,\infty)$ is invertible but $\exp : \mathbb R \to \mathbb R$ is not, since not every member of $\mathbb R$ has an image under the inverse function.  You might want to look with a fairly high-resolution microscope at how this particular author defines things.
