Counting Surjective Maps from a Finite Set to Itself

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The discussion centers on determining the number of surjective maps from a finite set S = {1,2,3,...,n} to itself. The correct answer is n!, as each element must map uniquely to cover all elements in the range. The confusion arises from the assumption that surjective maps can be counted as n^n, which does not account for the requirement of a one-to-one mapping for surjectivity in finite sets. It is clarified that without a one-to-one mapping, the function cannot be onto, thus reinforcing that n! is indeed the correct count of surjective maps. The conclusion emphasizes the necessity of one-to-one mapping in establishing surjectivity for finite sets.
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Homework Statement



Let S = {1,2,3,...,n}

How many surjective maps are there from S to S?

Homework Equations



n/a

The Attempt at a Solution



The book's answer is n!

However, I thought that total number of surjective maps = n^n because 1-1 isn't required. Where am I wrong?
 
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This is understandable as a surjective map from itself to itself, and this is essential just the set S being rearranged, so the answer will be n!.

Mat
 
1-1 is implicitly required, because each element of the domain can only map to one element of the range - so to map to every element of the range you'll need a 1-1 mapping.
 
battousai said:

Homework Statement



Let S = {1,2,3,...,n}

How many surjective maps are there from S to S?The book's answer is n!

However, I thought that total number of surjective maps = n^n because 1-1 isn't required. Where am I wrong?

But since S is finite, if the map wasn't 1-1 it couldn't be onto, could it?
 
Arianwen said:
1-1 is implicitly required, because each element of the domain can only map to one element of the range - so to map to every element of the range you'll need a 1-1 mapping.

LCKurtz said:
But since S is finite, if the map wasn't 1-1 it couldn't be onto, could it?

Got it. Thanks.
 
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