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In the first chapter of an introductory topology text, I got stuck at the following exercise.

Define $f:A \to B$. By $f^{-1}$ we denote the preimage operation.

$$\text{If for any $X\subset A$, we have $f^{-1}\big[f[X]\big] = X.$ Then $f$ is injective.} $$ Pick $y,y' \in f[X]$ such that $y = y'$.

By assumption we can find $x,x' \in X$ such that $y = f(x)$ and $y' = f(x')$.

To prove that $f$ is injective we show that $x = x'$.

By definition we have \begin{align*} x &= f^{-1}(y) \\ &= f^{-1}(y') \\ &= x' \ . \end{align*}

I do not seem to be using the arbitraryness of $X$. It looks like this argument would work even if we just have inclusion of $X$ in $f^{-1}\big[f[X]\big]$.

Git Gud
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Mussé Redi
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3 Answers3

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You want to show that the preimage consists of at most one point.

Suppose $f(x_1) = f(x_2)$, and let $X=\{x_1\}$.

Then $\{x_1\} = f^{-1} (f ( \{x_1\} ) )$ and since $x_2 \in f^{-1} (f ( \{x_1\} ) )$, this implies $x_2 = x_1$, hence $f$ is injective.

copper.hat
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  • Isn't it suggestive to make an assumption for $X$? – Mussé Redi Mar 09 '15 at 17:56
  • I don't understand your question. – copper.hat Mar 09 '15 at 17:59
  • We should prove it for arbitrary $X$. Not only for singleton sets. – Mussé Redi Mar 09 '15 at 18:00
  • I think you misunderstand your question. It is asking you to prove that $f$ is injective if it satisfies the property $f^{-1}(f(X)) = X$ for any $X \subset A$. You are not proving something about $X$, you are proving something about $f$. – copper.hat Mar 09 '15 at 18:02
  • You are trying to prove that if $f(x_1) = f(x_2)$ then you must have $x_1=x_2$. This is what injective means. – copper.hat Mar 09 '15 at 18:03
  • As an aside, all the other answers do exactly what I have done above. – copper.hat Mar 09 '15 at 18:06
  • My point is that your $X$, does not seem arbitrary but our assumption says that ot should be. However I am aware of the fact that we are proving a property about $f$, not $X$. – Mussé Redi Mar 09 '15 at 18:20
  • You are missing the point. You are given a property the the function $f$ satisfies. We use that property to show that it is injective. The property says that for any $X \subset A$ that we choose we have blah, blah, blah. So, I can choose any specific $X$ I want to prove the property of injectiveness. – copper.hat Mar 09 '15 at 18:25
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    @MusséRedi: In fact, saying that $X=f^{-1}(f(X))$ for any SINGLETON X would be enough to conclude that $f$ is injective, so we don't need to much arbitraryness. – sranthrop Mar 09 '15 at 18:26
  • Eventually we use a singleton to prove injectiveness, as shown in @sranthrop's answer. So you could argue that any singleton will do. I'm just not sure that we can fix $X$, since the statement calls for arbitrary $X$. – Mussé Redi Mar 09 '15 at 22:02
  • Correction. @copper.hat's second comment makes sense. Indeed, we could just prove it for any $X$. – Mussé Redi Mar 09 '15 at 22:08
  • @MusséRedi: Every single answer here, including the one you accepted, does exactly what I have done above. – copper.hat Mar 09 '15 at 22:11
  • My confusion about the arbitraryness has cleared up. Indeed, we all use the same argument. – Mussé Redi Mar 09 '15 at 22:25
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If $f$ isn't injective you can't say that $x = f^{-1}(y)$. The best you have is $x \in f^{-1}(y)$ since the latter object is a set.

Try this. Your stated hypothesis is that $f^{-1}[f[\{x\}] = \{x\}$.

If $f(x') = f(x)$, then $f(x') \in f[\{x\}]$ so that $x' \in f^{-1}[f[\{x\}]$.

This means $x' \in \{x\}$, implying $x = x'$.

Umberto P.
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Note that in general $f^{-1}(\{y\})$ is a set and can contain more than one element. Example: $f:\mathbb R\to\mathbb R^2,\ x\mapsto x^2$. Then if $y=4$ we have $f^{-1}(\{y\})=f^{-1}(\{4\})=\{-2,2\}$.

In your case you can use $X=f^{-1}(f(X))$ for each subset $X$. Now, if $f(x)=f(x')$, then $\{x\}=f^{-1}(f(\{x\}))=f^{-1}(\{f(x)\})=f^{-1}(\{f(x')\})=f^{-1}(f(\{x'\}))=\{x'\}$ and hence $x=x'$.

sranthrop
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