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The full problem I'm working on requires me to show 3 statements are equivalent; however, I am only having an issue with showing one particular implication. I want to make clear I am not looking for an answer, I simply want to know where my issue is and what information I should look into to better my understanding of the problem or some point in the right direction.

Problem. For $f : X \to Y$, prove that

$$ \text{$f$ is injective} \quad \Rightarrow \quad \forall y \in Y, \ \lvert f^{-1}(\{y\}) \rvert \leq 1 $$

Attempt at Solution:

I know that $y \in Y$, $\exists x \in X$ s.t. $y=f(x)$. So, $$\lvert f^{-1}(\{y\})\rvert \leq 1 \quad \Leftrightarrow \quad \lvert f^{-1}(\{f(x)\}) \rvert \leq 1,$$ where $x \in X$.

I believe applying a function to a set is equivalent to applying that function to the set's elements. This is where I assume my issue is. However, my professors apply this reasoning often, so I can't help but think it's true.

$$ \Leftrightarrow \quad \lvert f^{-1}(f(\{x\})) \rvert \leq 1 $$

I know that "$f$ is injective" $\Leftrightarrow$ $\forall A \subseteq X$, $f^{-1}(f(A))=A$,

$$\lvert\{x\}\rvert \leq 1$$

I'm interpreting this to mean the $\forall y \in Y$, $-1 \leq x \leq 1$ where $x=f^{-1}(y)$ (which is clearly wrong). I also feel this step is a leap, but I'm not to sure on how to interpret a set containing exclusively $x$ to be different than $x$ itself.

Sangchul Lee
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Noaline
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  • Also, be careful with the meaning of $|{x}| \le 1$. This means that the number of elements of set ${x}$ is $\le 1$, which is true. Actually, you know it is precisely $1$. However, you cannot say anything about the nature of $x$. You don't know if $X$ and $Y$ are subsets of $\mathbb{R}$ or some other set. – D. Ungaretti Aug 29 '19 at 22:22
  • @Daniel Oh, so the conclusion holds true? That is great to know as I assume this will appear again. Thank you for this clarification. – Noaline Aug 29 '19 at 23:29

1 Answers1

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"I know that$y \in Y, \exists x \in X$such that$y=f(x)$" is false unless $f$ is surjective, which is not given. Try this when $X$ has one element and $Y$ has infinitely many.

What you know is $y \in f(X), \exists x \in X$ such that $y = f(x)$. That is, every element of the image of $f$ has a pre-image. But you only know this for the image of $f$, which need not be all of $Y$.

Eric Towers
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  • So would it be reasonable to state that f(X)⊆Y, and that what I stated is wrong because f may not map X to all of Y? – Noaline Aug 29 '19 at 23:33
  • Yes. Since $f$ is not surjective (synonym: onto), the image of $f$ need not be all of $Y$. – Eric Towers Aug 30 '19 at 01:03