Prove that the function $f$ is one-to-one if and only if $f^{-1}(y)$ contains at most one element for every $y \in Y$.

Question

Let $f$ be a function from $X$ to $Y$

(Note: if this is a duplicate please tell me how you searched to find the original question. Thank you).

Googling "proving injective function preimage" gives us the following question which is very similar. — Noble Mushtak, Oct 23 '16 at 01:29
It's utterly obvious from the definition. A "formal proof" of this statement makes it certainly less believable. — Christian Blatter, Oct 23 '16 at 08:51
@ChristianBlatter I like the the sentiment in that last statement but I don't think my college professors would let me write that in on a test. :-) — Idle Fool, Oct 23 '16 at 14:08

score 1 · Answer 1 · answered Oct 23 '16 at 01:53

Proof:

($\Rightarrow$) If $f$ are (one-to-one)injective this means that $f(x_1)=f(x_2)$ implies $x_1=x_2$. Let $x_1,x_2\in f^{-1}(\left\{y\right\})$ with $y\in Y$ so $f(x_1)=y=f(x_2)$ then by the hypotesis $x_1=x_2$ and your claim is true.

($\Leftarrow$) Suppose that $f^{-1}(\left\{y\right\})=\left\{x\right\}$ for all $y\in Y$. Thus, if you choose $x_1$ and $x_2$ with $f(x_1)=y=f(x_2)$ then, $x_1,x_2\in f^{-1}(\{y\})=\{x\}$ therefore $x_1=x_2=x$. So $f$ is injective.

Prove that the function $f$ is one-to-one if and only if $f^{-1}(y)$ contains at most one element for every $y \in Y$.

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