Injectivity and Surjectivity of the Exponential Function

Question

Why is the exponential function injective but not surjective?

Answer 1

The real valued function $f=\exp:\mathbb R\to \mathbb R$ has the properties that $f(0)=1$, $f'=f$ and $f(x+y)=f(x)f(y)$ for all $x,y\in \mathbb R$. Thus, $1=f(0)=f(x+(-x))=f(x)f(-x)$ and in particular, $f(x)\ne 0$ for all $x\in \mathbb R$. So, $f$ is not surjective. Since $f$ is continuous, it thus also follows from the intermediate value theorem that $f$ either attains only positive values or only negative values. As $f(0)=1$, it follows that $f(x)>0$ for all $x\in \mathbb R$. Now, it follows that $f'(x)=f(x)>0$, and thus $f$ is strictly increasing in $\mathbb R$. Every strictly increasing function is injective, thus $f$ is injective.

Interestingly, the exponential function can be extended to $\exp:\mathbb C\to \mathbb C$, where the function is no longer injective, and attains all complex values except for the sole exception of $0$.

Answer 2

Whether it's surjective depends on what you consider the codomain of the function. Remember, if you have a function $f:A\to B$, then the set $A$ is called the domain of the function and $B$ is called the codomain. $f$ is surjective if and only if $f(A)=B$ where $f(A)=\{f(x)\mid x\in A\}$, i.e. $f$ applied to all points of the domain hits all points of the codomain. Now, if you consider $\exp:\mathbb R\to\mathbb R$ then it's clearly not surjective, because it only hits the positive real numbers. If you consider $\exp:\mathbb R\to\mathbb R^{>0}$, then given that codomain, $\exp$ would be surjective, because $\exp(\mathbb R)=\mathbb R^{>0}$.

A function $f:A\to B$ is injective, iff $f(x)=f(y)\Rightarrow x=y$ for all $x,y\in A$. This is the case for the exponential function, since it is strictly monotone.

In general both properties (injectivity and surjectivity) are totally unrelated (though there are some exemptions to the rule).