Let $f: A \rightarrow B$, and let $C$ be a subset of A. Then it can be shown that it is the case that $C \subseteq f^{-1}[f(C)]$. However, what I don't understand is why equality between the left-hand side and right-hand side doesn't hold. This is the explanation given in my book:

Given $f: A \rightarrow B$, suppose $C_1, C_2$ are nonempty subsets of A such that $C_1 \cap C_2 = \varnothing$ and $f(C_1)=f(C_2)$. Then $f^{-1}[f(C_1)]=C_1 \cup C_2$, and this is larger than $C_1$. This shows that equality does not hold, i.e. that $C \neq f^{-1}[f(C)]$.

I don't understand the above justification - how does the author conclude $f^{-1}[f(C_1)]=C_1 \cup C_2$ from the assumptions?

Edit: the intent behind this question is clearly not to prove the book wrong, but to prove why equality doesn't hold between $C$ and $f^{-1}[f(C)]$. Please show me that instead.

  • $\begingroup$ What happen if $f: \mathbb R \Rightarrow \mathbb R$ is constant? $\endgroup$
    – Mr. T
    Nov 2, 2017 at 15:24
  • $\begingroup$ Is there an additional assumption that $f$ is not injective? $\endgroup$ Nov 2, 2017 at 15:25
  • $\begingroup$ @yinnonsanders No, this is all there is to it. No further context that could possibly be relevant to this particular section is omitted. It is not assumed that $f$ is surjective, injective or bijective. Nothing is assumed about the nature of A, or B, or indeed any other sets involved in the assumptions. $\endgroup$
    – Ius Klesar
    Nov 2, 2017 at 15:29
  • $\begingroup$ @yinnonsanders it is implied that $f$ is noninjective, since $f(C_1)=f(C_2)$ where the two sets in question are nonempty and disjoint $\endgroup$ Nov 2, 2017 at 15:56
  • $\begingroup$ @AndresMejia Exactly, two such subsets exist iff $f$ is noninjective. $\endgroup$ Nov 2, 2017 at 16:08

2 Answers 2


A couple of examples that might make you understand why equality doesn't necessarily hold:

  1. $f(x) = x^2$ with $C = \{1\}$. We have $f(C) = \{1\}$ but $f^{-1}(\{1\}) = \{ -1, 1 \} \supset C.$

  2. $f = \cos$ with $C = [0, \pi]$. We have $f(C) = [-1, 1]$ but $f^{-1}([-1, 1]) = \mathbb R \supset C.$

  • $\begingroup$ Thank you, that was actually very intuitive! But then, how would I prove it? And could you help me understand the proof provided in my book? $\endgroup$
    – Ius Klesar
    Nov 2, 2017 at 15:49
  • $\begingroup$ To prove that an equality does not hold, an example like one of those I gave is just fine. There's no need for a formalized proof. $\endgroup$
    – md2perpe
    Nov 2, 2017 at 16:32

This is not true. Let $f:\mathbb R \to \mathbb R$ be the map $x \mapsto 0$.

Let $C_1,C_2=\{1\},\{2\}$ respectively.

Then all of the conditions are satisfied, but $f^{-1}f(C_1)=f^{-1}(\{0\})=\mathbb R$.

The most we can conclude is that $C_1 \cup C_2 \subset f^{-1}f(C_1)$.

  • $\begingroup$ So my book (Steven Lay's "Analysis with an Introduction to Proof" got it wrong? $\endgroup$
    – Ius Klesar
    Nov 2, 2017 at 15:32
  • $\begingroup$ I'm sorry, but I'm really not following this at all. Where did you get the $C$ from? There's no $C$ involved in Lay's explanation, how do you know that $C \subsetneq C_1 \cup C_2$? And could you somehow edit your comment with a proof of the fact that inequality doesn't hold, if Lay's proof isn't correct? $\endgroup$
    – Ius Klesar
    Nov 2, 2017 at 16:02
  • $\begingroup$ But $C_1 \cup C_2 \subset f^{-1}[f(C_1)]$ is not included in his proof anywhere, what do you mean by replacing it with inclusion? And isn't inclusion already in $C_1 \cup C_2 \subset f^{-1}[f(C_1)]$? So what is there to replace? $\endgroup$
    – Ius Klesar
    Nov 2, 2017 at 16:10
  • $\begingroup$ He concludes that they are equal. I'm saying to replace that equality with inclusion. I deleted comments since this was getting quite long. $\endgroup$ Nov 2, 2017 at 16:12

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