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  1. Let $X$, $Y$ be sets and let $f:X \to Y$ be a map. For any subset $C \subseteq Y$ of the codomain $Y$, we consider its preimage $f^{−1}(C) \subseteq X$ in the domain $X$: $f^{−1}(C) := \{x \in X : f(x) \in C\}$.

(Note: the map $f$ here is not assumed to be invertible; thus the notation $f^{−1}$ here denotes “preimage”, not the inverse of $f$ .)

Show that $f$ is injective if and only if for any subset $C \subseteq Y$ such that $C$ is empty or a singleton, its preimage $f^{−1}(C) \subseteq X$ is empty or a singleton.

so if $C$ is empty or a singleton, the element will only hit at most one target on the range. because whatever we ask of the empty set will be true, so what are the steps i need to show in order to prove that it is injective. (injectivity, F(x1) = F(x2) in Y. right?

balddraz
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user271716
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2 Answers2

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Note that under your assumptions, for each $p\in X$, you have: $$p\in f^{-1}(f(p)).$$ That is $f^{-1}(f(p))=\{p\}$

Then, if $f(p)=f(q)$ necessarily $$\{p\}=f^{-1}(f(p))=f^{-1}(f(q))=\{q\}$$ so, $p=q$.

janmarqz
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Hint:

To prove injectivity of $f$ it is enough to prove that $f(x_1)=f(x_2)$ implies $x_1=x_2$.

If $f(x_1)=y=f(x_2)$ have a look at $f^{-1}(\{y\})$ wich is the preimage of a singleton.

edit (on request):

Let it be that $f^{-1}(C)$ is empty or is a singleton whenever $C\subseteq Y$ is a singleton. To be shown is that $f$ is injective. Let it be that $x_1,x_2\in X$ with $f(x_1)=f(x_2)$. Set $C:=\{f(x_1)\}$ is a singleton so we are allowed to conclude that $f^{-1}(C)$ is empty or a singleton. The elements $x_1,x_2$ both belong to this set so it is not empty. Then it is a singleton and $f^{-1}(C)=\{x_1\}=\{x_2\}$ so that $x_1=x_2$. Proved is now that $f$ is injective.

drhab
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  • So the preimage implies, the elements in the domain would now be the elements of the range, and then the ones in the codomain are the elements of the original domain? am i right to say that? – user271716 Sep 22 '15 at 13:46
  • The preimage mentioned in my answer must be empty of must be a singleton. It is not empty (e.g. it contains $x_1$) so it must be a singleton, and one that contains both $x_1$ and $x_2$ as elements. What can you conclude about $x_1$ and $x_2$? – drhab Sep 22 '15 at 13:50
  • we can conclude that x1=x2? so that it maintains its injectivity? – user271716 Sep 22 '15 at 13:55
  • It is a proof that $f$ is injective. Wasn't that your aim? – drhab Sep 22 '15 at 14:01
  • is it okay if i have a detailed proof of how to do? because i dont know the steps to go about doing it. i know the concept, but i dont know how to show it... – user271716 Sep 22 '15 at 14:31