MHB What Comparison Sign To Assert f^(-1)(f(A))? A True?

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For a function f from set X to set Y and subset A of X, the relationship f^(-1)(f(A)) can be compared to A using the subset relation. It is established that A is always a subset of f^(-1)(f(A)), meaning A ⊆ f^(-1)(f(A)). However, equality (A = f^(-1)(f(A))) is not guaranteed, as there may exist elements outside A that map to the same output in f(A). An example is provided with constant functions, where f^(-1)(f(A)) equals X regardless of A. Thus, the comparison sign can be either ⊆ or =, depending on the nature of the function.
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Let f be a function from a set X to a set Y, moreover, A ⊆ X. What comparison sign canput instead? to assert "f ^(−1) (f(A)) ? A" become true? (Possible signs of comparison in this : ⊆, ⊇, =. It is necessary to take into account all options.

f ^(−1) - inverse of fall options.), Let f be a function from a set X to a set Y, moreover, A ⊆ X. What comparison sign canput instead? to asserte-one(f (a))?become true? (Possible signs of comparison in this and the following two problems: ⊆, ⊇, =. It is necessary to take into accountall options.)
 
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ranga519 said:
Let f be a function from a set X to a set Y, moreover, A ⊆ X. What comparison sign canput instead? to assert "f ^(−1) (f(A)) ? A" become true? (Possible signs of comparison in this : ⊆, ⊇, =. It is necessary to take into account all options.


Suppose $x\in A$. Then $f(x)\in f(A)$ so that $x\in f^{-1}(f(A))$. But there might exist $y\notin A$ such that $f(y)\in f(A)$. That y would also be in $f^{-1}(f(A))$. So what we can say is that $A\subseteq f^{-1}(f(A))$ with the "=" possible but not necessarily. For example if f is a "constant function", $f(x)= y\in Y$ for all $x\in X$ where y is a specific member of Y, then $f^{-1}(f(A))= X$ for A any subset of X.
 
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