# Why $f^{-1}(f(A)) \not= A$ [duplicate]

This question already has an answer here:

Let $A$ be a subset of the domain of a function $f$. Why $f^{-1}(f(A)) \not= A$. I was not able to find a function $f$ which satisfies the above equation. Can you give an example or hint. I was asking for an example function which is not addressed here

## marked as duplicate by Martin Sleziak, Asaf Karagila♦, Amzoti, vonbrand, achille huiApr 22 '13 at 14:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• $A$ is not "a subset of a function", but a subset of its domain. – Michael Greinecker Apr 22 '13 at 7:27
• @MichaelGreinecker. Corrected. – Vinod Apr 22 '13 at 8:06
• @Duplicate. When I was asking the question, no relevant suggestions came up.Not an exact duplicate. – Vinod Apr 24 '13 at 6:24
• I am fairly sure that my answer to the currently duplicate link contains an example. – Asaf Karagila Apr 24 '13 at 6:57

## 4 Answers

Any noninjective function provides a counterexample. To be more specific, let $X$ be any set with at least two elements, $Y$ any nonempty set, $u$ in $X$, $v$ in $Y$, and $f:X\to Y$ defined by $f(x)=v$ for every $x$ in $X$. Then $A=\{u\}\subset X$ is such that $f(A)=\{v\}$ hence $f^{-1}(f(A))=X\ne A$.

In general, for $A\subset X$, $A\subset f^{-1}(f(A))$ but the other inclusion may fail except when $f$ is injective.

Another example: define $f:\mathbb R\to\mathbb R$ by $f(x)=x^2$ for every $x$. Then, $f^{-1}(f(A))=A\cup(-A)$ for every $A\subset\mathbb R$. For example, $A=[1,2]$ yields $f^{-1}(f(A))=[-2,-1]\cup[1,2]$.

Let $f:\mathbb{R}\to\mathbb{R}$ be given by $f(x)=1$ for $x\geq 0$ and $f(x)=0$ for $x<0$ and $A=[0,1]$, then $$f(A)=\{1\}\;\;\Longrightarrow \;\;f^{-1}(f(A))=f^{-1}(\{1\})=[0,\infty)\neq [0,1]=A.$$

First, you mean, for a function $f:X\to Y$ and a subset $A\subset X$, why is it that $f^{-1}(f(A))\ne A$. Note that in order to describe a function you must be clear about its domain and codomain.

Now, try this: $X=Y=\mathbb N$ and $f:X\to Y$ given by $f(n)=1$. Then for the set $A=\{1\}$ compute $f^{-1}(f(A))$.

If $f:X\longrightarrow Y$ is not injective and $A\subset X$. Then take $x \in X-A$ such that $f(x)=f(a)$ for some $a\in A$ then we have that $x\in f^{-1}(f(A))$ although $x\not \in A$.

• We cannot use arbitrary $A\subset X$. E.g. let $f:\mathbb R\to\mathbb R$ be given by $f(x)=x$ except at zero where $f(0)=1$. Then $A=\{-1\}$ does not work. – Karl Kronenfeld Apr 22 '13 at 7:20