Why is the exponential function injective but not surjective?

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Layne Murillo · Accepted Answer

The real valued function   f  =  exp  :      R    →      R   has the properties that f(0)=1, f′=f and f(x+y)=f(x)f(y) for all   x  ,  y  ∈      R  . Thus, 1=f(0)=f(x+(−x))=f(x)f(−x) and in particular,   f  (  x  )  ≠  0 for all   x  ∈      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)&amp;gt;0 for all   x  ∈      R  . Now, it follows that f′(x)=f(x)&amp;gt;0, and thus f is strictly increasing in       R  . Every strictly increasing function is injective, thus f is injective.Interestingly, the exponential function can be extended to   exp  :      C    →      C  , where the function is no longer injective, and attains all complex values except for the sole exception of 0.

Why is the exponential function injective but not surjective?

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