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Suppose that we have a function $f$ $:$ $X$ $\to$ $Y$, with

$ X $ $=$ {1,2,3,4,5,6}
$Y$ $=$ {a,b}

define $$ f(x) = \left\{ \begin{array}{ll} a & \mbox{if $x =1$ OR $2 $ OR 3} \\ b & \mbox{if $ x = 4 $ OR 5 OR 6} \ \end{array}\right. $$ my question is, when we take the f inverse of {a,b} will we get {1,2,3,4,5,6} OR {{1,2,3},{4,5,6}} ?

Edited: I also have another doubt, do $f^{-1}(a)$ and $f^{-1}$({$a$}) mean the same?

Rohan Marwah
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2 Answers2

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The preimage or inverse image of the set $Y=\{a,b\}$

is the subset of $X$ defined by $f^{-1}(Y)=\{x\in X|f(x)\in Y$},

which in this case is $\{1,2,3,4,5,6\}=X$.

J. W. Tanner
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Let $A:=${$a$}; $B:=${$b$}; $A \cup B=${$a,b$}.

$f^{-1}(A)=f^{-1}$({$a$})$=$

{$1,2,3$}$\subset ${$1,2,3,4,5,6$}.

$f^{-1}(B)=f^{-1}$({$b$})$=$

{$4,5,6$}$\subset ${$1,2,3,4,5,6$}.

$f^{-1}(A \cup B)=$

$f^{-1}(${$a,b$}$)=f^{-1}(A)\cup f^{-1}(B)=$

$f^{-1}(${$a$}$)\cup f^{-1}(${$b$}$)=$

{$1,2,3,4,5,6$}.

Used: $f^{-1}(A \cup B)=f^{-1}(A)\cup f^{-1}(B)$.

Peter Szilas
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  • Thanks. could you please address the second part of the question as well. – Rohan Marwah Nov 24 '19 at 18:12
  • Cosmic.$f^{-1}(A)=f^{-1}(${a}$)$ is the inverse image (a set) of the set A. If $f$ were one to one $f^1(a)=x$ one element of the domain $X$.Makes sense? – Peter Szilas Nov 24 '19 at 18:24
  • can't we define the inverse image of an element 'a' in the above (many-one) mapping given in the question? – Rohan Marwah Nov 24 '19 at 18:30
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    I would use $f^{-1}(a)=x$, if $f$ is one to one , both $a,x$ are elements. $f^{-1}$ is the inverse function of $f$. $f^{-1}(${$a$}$)= D$, where {$a$},$D$ are sets for many to one.See https://en.m.wikipedia.org/wiki/Image_(mathematics)#Inverse_image – Peter Szilas Nov 24 '19 at 18:44