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monomorphisms of category of sets
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(Theorem)
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Proof. Assume
 is a monomorphism. Then, by definition of monomorphism, given any two maps
 , if
 , then  . Suppose  and  are two elements of  such that
 . Let  be a set with one element, let  be the map which sends this one element to  and let  be the map which sends this one element to  . Because
 , we have
 . Since  is a monomorphism,  , so  . This implies that  is injective. 
Proof. Assume
 is injection. If  is empty, the result is trivial, so we assume that  is not empty; let  be an element of  . Set
We claim that  is a function from  into  . Suppose that  is an element of  . If
 for any  , then we have exactly one element of  with  as the first element, namely  . If  for some  , then we the pair  with  as first element; were there another pair with  as first element, then we would have
 but, as  is an injection,
 would imply  , so this would not be a distinct pair. Hence  is a function. Furthermore, by construction
 for all  , so  is a split monomorphism. 
Note that the second theorem is stronger than a simple converse to the first theorem -- it states that an injection is not just a monomorphism, but that it is actually a split monomorphism. In particular, this means that, in the category of sets, all monomorphisms are actually split monomorphisms.
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"monomorphisms of category of sets" is owned by rspuzio. [ full author list (2) ]
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(view preamble)
Cross-references: converse, simple, split monomorphism, implies, maps, injection, category of sets, monomorphism
There is 1 reference to this entry.
This is version 4 of monomorphisms of category of sets, born on 2007-02-18, modified 2008-04-30.
Object id is 8927, canonical name is MonomorphismsOfCategoryOfSets.
Accessed 503 times total.
Classification:
| AMS MSC: | 18-00 (Category theory; homological algebra :: General reference works ) |
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Pending Errata and Addenda
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