Prove that $\exp: \mathbb{R} \mapsto (0,\infty)$ is a bijection.
Okay, so the first part is really easy: injectivity follows directly from writing the exponential function as a series.
Surjectivity… is not so easy. I have no idea how to show it directly, and I was thinking maybe I could show that
$\inf(e^x) = 0$ and $\sup(e^x) = \infty$, and that $e^x$ is continuous over all of $\mathbb{R}$. Does this imply surjectivity over $(0,\infty)$?
Best Answer
Injectivity
If you know that $\exp' = \exp$, use the fact that $\exp'(x) = \exp > 0 \;\; \forall \, x$.
Alternatively, if you know that $\exp (a + b) = \exp a \exp b$, then assume $\exp x = \exp y$. Conclude $\exp(x - y) = 1$. Assume WLOG $x \ge y$, and then use the series expansion of $\exp(x - y)$ to show $x - y = 0$, hence $x = y$.
Surjectivity
Fix $x \in (0, \infty)$. Find $a$ and $b$ with $e^a < x$, $e^b > x$. Apply Intermediate Value Theorem to the interval between $a$ and $b$.