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Is $F:R\to[0, ∞)$ where $F(x) = e^x$ a bijection?

For a function to be surjective, the function must hit all elements belonging to the CODOMAIN (Which is $[0,∞)$ right?) or does it simply have to hit all possible values of elements that can be received after plugging in any value from the domain.

I think as I'm writing this out instead of trying to figure it out in my mind it's making more sense.

A function is surjective if, for any value of the domain(R) substituted into the function, a value of $[0,∞)$ must be the result am I correct, and ALL VALUES MUST BE HIT.

Sorry about the stream-of-though-esque articulation.. :(

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    $\begingroup$ It is a bijection from $\Bbb R$ onto $(0, \infty)$ $\endgroup$ – Ishfaaq Aug 14 '15 at 6:33
  • $\begingroup$ You've got surjective backwards: you need to be able to pick any $y$ from your range, $[0,\infty]$, and find some $x$ in the domain that gets mapped to $y$. What you wrote is basically the definition of range/codomain. $\endgroup$ – pjs36 Aug 14 '15 at 6:35
  • $\begingroup$ "does it simply have to hit all possible values of elements that can be received after plugging in any value from the domain": All the functions satisfy that. Think about it... $\endgroup$ – Najib Idrissi Aug 14 '15 at 7:22
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You are correct with your original definition of a bijection. A bijection means that every single point in the codomain must be hit. Now I ask you, since $0$ is in the codomain, is it ever hit?

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  • $\begingroup$ didn't notice that! thank you! $\endgroup$ – Eddard Aug 14 '15 at 6:39
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$e^x$ is in bijective correspondence with $(0, \infty)$ but not $[0,\infty)$. The inverse is $\ln(x)$ on the former interval. The latter interval, you see it fails to be surjective since $e^x > 0$ for all $x\in \mathbb{R}$.

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  • $\begingroup$ I see this not, thank you! $\endgroup$ – Eddard Aug 14 '15 at 6:39
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It is not a bijection. It is first 1-1 function since $e^x=e^y \iff e^{x-y} = 1 \iff x-y=0\iff x=y$. It is not onto since $0$ has no preimage.

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It’s not surjective: $e^x$ is a positive real number, so $0$ and $\infty$ aren’t elements of the image. But you can prove the following: $$\lim_{x \to -\infty} e^x = 0 \qquad \lim_{x \to \infty} = \infty$$ The exponential function is continuous, so by intermediate value theorem, $x \mapsto e^x: \mathbb{R} \to (0,\infty)$ is surjective. It is injective, since it is strictly monotonic increasing.

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Given $f:\mathbb{R}\Rightarrow \left(0,\infty\right)$ and $f(x) = e^x$

Here Function $f(x)$ is defined for all real $x$

and Now Condition for Checking one to -one function

For that Let $x_{1},x_{2}\in \mathbb{R}$ (Domain)

If $f(x_{1}) = f(x_{2})\Rightarrow e^{x_{1}} = e^{x_{2}}\Rightarrow x_{1}=x_{2}$

So function $f(x)$ is one to one

and Condition for Onto

Let $y = f(x) = e^x\Rightarrow e^x=y$

taking $\ln$ on both side, We get $x=\ln(y)\;,$ Which is defined when $y>0$

So we get Range $y\in \left(0,\infty\right) = $ Codomain

So Function $f(x)$ is one-to -one and onto , So it is Bijective function.

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