Is there a finitely complete category with terminal object but NO subobject classifier? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T20:51:55Z http://mathoverflow.net/feeds/question/14815 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14815/is-there-a-finitely-complete-category-with-terminal-object-but-no-subobject-class Is there a finitely complete category with terminal object but NO subobject classifier? B. Bischof 2010-02-09T21:41:51Z 2010-02-10T13:10:32Z <p>This came up today while thinking about topoi in seminar, as the title suggests my question is;</p> <blockquote> <p>Is there a finitely complete category with terminal object but NO subobject classifier?</p> </blockquote> <p>Hopefully if the answer is yes, you can give an interesting example that is useful in some way. One of my professors has a saying that </p> <blockquote> <p>for every definition you should have an interesting example and an interesting non-example</p> </blockquote> <p>So I am trying to stick to this. </p> <p>Thanks!</p> http://mathoverflow.net/questions/14815/is-there-a-finitely-complete-category-with-terminal-object-but-no-subobject-class/14820#14820 Answer by Steven Gubkin for Is there a finitely complete category with terminal object but NO subobject classifier? Steven Gubkin 2010-02-09T22:15:51Z 2010-02-10T13:10:32Z <p>I am pretty sure I proved that the category of groups has no subobject classifier at some point. I will try and edit this post with the proof when I have more time, but I think this is an example for you to think about.</p> <p>EDIT: Ya, this isn't too bad. If there was a subobject classifier $\Omega$ in Groups, then by looking at the characteristic map of the injection of $0 \rightarrow A$ for each map, you will see by writing out the diagrams that A will have to inject into $\Omega$. But then $\Omega$ is bigger than every cardinal since there is a group of every cardinality. That doesn't fly. </p> <p>EDIT of EDIT: I guess I should flesh this out in greater detail.</p> <p>The first thing to note is that the category of groups has a zero object (it's terminal object is also initial). I will write 0 for this.</p> <p>Say Groups had a subobject classifier $0 \stackrel{true}\longrightarrow \Omega$. Note that the map true must be the unique map out of 0. Also note that the unique map $0 \stackrel{!}\rightarrow A$ is a monomorphism (i.e. injection) for every A. Thus </p> <pre><code> 0-----&gt;0 | | ! | | true | | \/ \/ A ----&gt;Ω </code></pre> <p>is a pullback square, where the lower map, $\chi$, is the characteristic map of !. I claim that $\chi$ is a monomorphism. This is because the ker($\chi$) maps to both A and 0 to make the diagram commute, so the inclusion of ker($\chi$) into A factors through 0 by the definition of a pullback. In other words, the kernel is trivial, so $\chi$ is an injection. Thus every group A admits an injection to $\Omega$ which is bad for set theoretic reasons.</p> http://mathoverflow.net/questions/14815/is-there-a-finitely-complete-category-with-terminal-object-but-no-subobject-class/14822#14822 Answer by Sridhar Ramesh for Is there a finitely complete category with terminal object but NO subobject classifier? Sridhar Ramesh 2010-02-09T22:20:45Z 2010-02-09T22:34:17Z <p>Yes, there are many, many finitely complete categories with no subobject classifier. Indeed, I daresay <em>most</em> finitely complete categories lack subobject classifiers (for some notion of "most").</p> <p>For example, any non-trivial meet semilattice is one (every object admits a morphism into the subobject classifier classifying its identity subobject; however, in a preorder, morphisms are unique. Accordingly, in a preorder with a subobject classifier, every object has a unique subobject, and thus, if there is a terminal object, every object is isomorphic to it).</p> <p>(I second Yuan's nitpick; a finitely complete category automatically has a terminal object)</p> http://mathoverflow.net/questions/14815/is-there-a-finitely-complete-category-with-terminal-object-but-no-subobject-class/14824#14824 Answer by Charles Rezk for Is there a finitely complete category with terminal object but NO subobject classifier? Charles Rezk 2010-02-09T22:23:00Z 2010-02-09T22:23:00Z <p>I assume "finitely complete" means "has finite limits". If so, it seems to me that counterexamples are pretty common. Mine would be the category of abelian groups.</p> <p>A subobject classifier is a monomorphism $u:1\to \Omega$, where $1$ is the terminal object, such that for any subobject $S\to X$, there is a unique map $f_S:X\to \Omega$ such that $S=f^{-1}(1)$. </p> <p>Notice that this means that any automorphism $f:\Omega \to \Omega$ such that $f\circ u = u$ must be the identity map. So $\Omega$ has no nonidentity automorphisms compatible with $u$.</p> <p>In the category of abelian groups, the terminal object is also the initial object. So $\Omega$ would be an abelian group with no non-identity automorphisms. Since multiplication by $-1$ is always an automorphism, such a group must be a $Z/2$-vector space. Most of those have non-trivial automorphisms, and the ones which don't are clearly not subobject classifiers.</p> http://mathoverflow.net/questions/14815/is-there-a-finitely-complete-category-with-terminal-object-but-no-subobject-class/14829#14829 Answer by Tom Church for Is there a finitely complete category with terminal object but NO subobject classifier? Tom Church 2010-02-09T23:00:21Z 2010-02-10T01:22:06Z <p>You should basically never expect anything to have a subobject classifier -- it's a ridiculously strong condition. Some examples showing how quickly it fails to exist:</p> <ul> <li><p>The category of groups is complete. If this category had a subobject classifier &Omega;, every subgroup would be the kernel of a map to &Omega;. But not every subgroup is normal.</p></li> <li><p>The category of Hausdorff spaces is complete. If it had a subobject classifier &Omega;, every subspace would be the preimage of a point in &Omega;. But not every subspace is closed.</p></li> <li><p>The category of rings with identity is complete. The terminal object is the zero ring (with 0 = 1). But the zero ring has no maps to any nonzero ring.</p></li> <li><p>The category of compact Hausdorff spaces is complete. As above every closed subspace should be the preimage of a point $\ast$ in &Omega; under a <em>unique</em> map to &Omega;. For the subspace {1} in {1,2} to be classified by a unique map, $\Omega\setminus \ast$ must be a single point, so &Omega; is discrete on two points. But then there is only one map from $S^1$ to &Omega;, while it has uncountably many subobjects.</p></li> </ul> <p><hr /></p> <p>Edit: the opposite of any cocomplete category will (with probability 1) give examples as well.</p> <ul> <li><p>The category of groups is cocomplete. A subobject classifier in the opposite category would be a group &Omega; so that every surjection G -> H is a pushout of &Omega; -> 1 and &Omega; -> G. This is the amalgamated free product of G with the trivial group over &Omega;, that is the quotient of G by the image of &Omega;. But if &Omega; is a group surjecting to the kernel of every map, that surjection cannot be unique (just consider Z -> 1).</p></li> <li><p>The category of sets (also Hausdorff spaces) is cocomplete. In the opposite category &Omega; would be a set (space) so that every surjection X -> Y was the quotient of X by the image of a map &Omega; -> X; this would imply that at most one fiber of X -> Y is not just a single point.</p></li> <li><p>The category of commutative rings with identity is cocomplete. A subobject classifier for the opposite category (the category of affine schemes) would be an affine scheme &Omega; with a $\mathbb{Z}$-point Spec $\mathbb{Z}$ -> &Omega; so that any monomorphism Y -> X is the base change to $\mathbb{Z}$ of a unique map X -> &Omega;. This is ridiculous; note that Spec $\mathbb{Q}$ -> Spec $\mathbb{Z}$ is injective, and you'll never get Spec $\mathbb{Q}$ as the fiber product of two $\mathbb{Z}$-points.</p></li> <li><p>The partial order category of ordinals is cocomplete. The initial object is 0; but no nonzero ordinal maps to (is &le;) 0. Thus &Omega; would have to be 0, but then we would have for any &alpha; &ge; &beta; that &alpha; was the coproduct (supremum) of &beta; with 0, which is false whenever &alpha; &gt; &beta;.</p></li> </ul> http://mathoverflow.net/questions/14815/is-there-a-finitely-complete-category-with-terminal-object-but-no-subobject-class/14834#14834 Answer by Alex Hoffnung for Is there a finitely complete category with terminal object but NO subobject classifier? Alex Hoffnung 2010-02-10T00:44:32Z 2010-02-10T00:44:32Z <p>There are also cases where there is no subobject classifier, but there is a weak subobject classifier -- meaning it only classifies monomorphisms with a certain property. Various categories of smooth spaces, as well as simplicial complexes are examples.</p>