Is there a category with a subobject classifier but which is not finitely complete? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T11:32:29Z http://mathoverflow.net/feeds/question/55955 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55955/is-there-a-category-with-a-subobject-classifier-but-which-is-not-finitely-complet Is there a category with a subobject classifier but which is not finitely complete? beroal 2011-02-19T04:03:37Z 2011-02-19T05:30:16Z <p>This is a reverse of the question <a href="http://mathoverflow.net/questions/14815/is-there-a-finitely-complete-category-with-terminal-object-but-no-subobject-class" rel="nofollow">“Is there a finitely complete category with terminal object but NO subobject classifier?”</a> From <a href="http://arxiv.org/abs/1012.5647" rel="nofollow">“An informal introduction to topos theory”</a> by Tom Leinster I learned that there is 3 definitions of a subobject classifier in some category C:</p> <ol> <li>where we directly work with morphisms of C: for every monomorphism there exists a characteristic morphism etc.;</li> <li>a terminal object in the category of monomorphisms and pullback squares;</li> <li>the functor Sub is representable.</li> </ol> <p>In order to make sense of (3) we need Sub which is defined via pullbacks in C. (3) requires C to have pullbacks. But (1) and (2) do not, though they imply existence of the terminal object. Is there a category with a subobject classifier and which is not finitely complete? (AFAIK subobject classifier → terminal object → (have pullbacks ↔ have finite limits = is finitely complete).)</p> http://mathoverflow.net/questions/55955/is-there-a-category-with-a-subobject-classifier-but-which-is-not-finitely-complet/55963#55963 Answer by Todd Trimble for Is there a category with a subobject classifier but which is not finitely complete? Todd Trimble 2011-02-19T05:30:16Z 2011-02-19T05:30:16Z <p>Yes: take the full subcategory of \$Set\$ whose objects are sets of cardinality 2 or less. </p>