I've been reading up on injection, surjection and bijection, but I haven't been able to find any examples that can help me figure out the solutions for these two:

  1. Let A = {1, 2, 3} and B = {a, b, c, d}. Find one surjection from B to A.

  2. Prove/disprove that the following function is injection and surjection (bijection): $f: (0,1) \rightarrow (1, e), f(x) = e^x$

Any help is much appreciated :)

  1. A surjection from B to A is simply a function where every element in A is an output of the function. One such function could be defined by $f(a) = 1, f(b) = 2, f(c) = 3, f(d) = 1$.

  2. To prove that the function $f(x) = e^x$ is a bijection, we can find the inverse. $f^{-1}(x) = \ln x$ (which is defined for this particular domain).

  • $\begingroup$ Thanks! Is it really enough to just find the inverse function to prove it is bijection? $\endgroup$ – saremisona Apr 16 '17 at 11:25
  • $\begingroup$ Yep! The fact that a function is bijective is equivalent to the fact that it has an inverse. Don't forget to prove that it is an inverse though (show that $(f \circ f^{-1})(x) = (f^{-1} \circ f)(x) = x$. $\endgroup$ – hazelmort Apr 16 '17 at 11:30

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