Surjection: Is f^-1(X) Surjective? Why?

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If a function f: N→X is surjective, the inverse image f^-1(X) is not a function but rather a set of natural numbers. The discussion clarifies that f^-1(X) is only a function if f is bijective, meaning both injective and surjective. The confusion arises from interpreting f^-1(X) as a function rather than recognizing it as a set. The correct interpretation suggests that for any function f: N→X, f^-1(X) equals N, making the surjectivity of f irrelevant to this specific question. Overall, the key takeaway is that the nature of f^-1(X) as a set distinguishes it from being a function.
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If a function f: N-->X is surjective , is f^-1(X) (its inverse image) also surjective? If so, why?
 
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f^-1(X) isn't a function...

Anyways, have you looked at any examples?
 
Hurkyl meant: isn't a function unless f is one-one, i.e. injective and surjective, i.e. bijective. Was this a trick question from some problem set?
 
No, Chris, I don't believe that's what Hurkyl meant! I started to interpret f-1(X) as if it were f-1(x) and say "that's not necessarily a function", but f-1(X) is the "inverse image" of X. It's not a function for the very good reason that f-1(X) is a set of natural numbers.

I suspect that the correct question was "If f: N->X is surjective is f-1(X)= N?" If I understand what is meant by "f:N->X", then "surjective" is irrelevant. For ANY function f:N->X, that is, "to every point in N assigns a point in X", f-1(X)= N.
 

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