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I know this question has been done before (Surjectivity implies injectivity), but I'm not pleased with the answer, because, it defines a "function" $g$, such that if $f(z) = x$, then $g(x) = z$. But $g$ is a function, only if $f$ is injective, and that's what we are trying to prove, so you can't use that. In fact, if $f(x) = f(y) = r$ and $x \not = y$, then $g(r) = x$ y $g(r) = y$ and $g$ is not a function.

Now, I have proved that if $f$ is injective, then it is surjective.

Let X be a set of order $n$ and $f:X \rightarrow X$, a injective function. First I supposed $f$ is not injective, therefore, there exist a $z\in X$, such that $z\not = f(x) \ \forall x\in X$, in particular $z\not = f(z)$. Then, I defined the set $B= \{f(z), f^2(z), f^3(z), ... f^n(z)\}$ and showed it was, in fact, equal to X. But, we can notice that $z\not = f^i(z) \ \forall i \leq n$ then $z\not \in B = X$, and we have arrived to a contradiction, since we parted by saying that $z\in X$, therefore $f$ is surjective.

Now, how would one proceed to show the counterpart?

Pepe Pérez
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Use the fact that the domain and range have the same number of elements. Since every element in the range is hit by the function when we assume surjectivity and every element in the domain is sent to only one element in the range by the definition of a function, then what can we conclude regarding injectivity?

M_B
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  • I do know, intuitively, that if it weren’t injective, then two elements would hit the same element in the image, but since they have the same cardinality, there would be an element left behind and therefore the function wouldn’t be surjective. But I don’t know how to write that, or exhibit an element that wouldn’t be in the image given that two element arrive the same element. – Pepe Pérez Sep 23 '18 at 13:56