I need to solve the following exercise.

Let $X$ be a non empty set and $P_2$ the set whose elements are all the subsets of $X$ that have $1$ or $2$ elements.

We have $f : X^2 \rightarrow P_2$, $(x, y) \mapsto \{x, y\}$

Let $Y\in P_2$. Determine $f^{-1}(\left\{{Y}\right\})$ and find its cardinality depending on the cardinality of $Y$.

The only things that I've managed to figure out is that $f$ is surjective and that $\text{card}(Y) = 1$ or $\text{card}(Y) = 2$.


1 Answer 1


If $ card(Y)=1 $ then it means $Y$ has only one element $x$. Thus $f^{-1}(\left\{{Y}\right\})=\left\{{(x,x)}\right\} $.

If $ card(Y)=2 $ then $ Y=\left\{{x,y}\right\} $ for some $x,y \in$ X. Thus $f^{-1}(\left\{{Y}\right\})=\left\{{{(x,y)},{(y,x)}}\right\} $

  • 1
    $\begingroup$ Wouldn't $f^{-1}(Y)=\{(x,y);(y, x) \}$? $\endgroup$ Oct 9, 2021 at 16:19
  • $\begingroup$ Yes, indeed @intheshadow $\endgroup$ Oct 9, 2021 at 16:25

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