Zhen Lin

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20h	comment	Spheres in different dimension are not homotopy equivalent Well, the spheres (for $n > 1$) are all simply-connected, so there aren't any interesting covering spaces...
22h	comment	Does taking the direct limit of chain complexes commute with taking homology? I have a preprint on arXiv; you can find my email address therein.
22h	comment	Questions about epimorphisms and projectives in functor categories See this question.
23h	comment	Questions about epimorphisms and projectives in functor categories I meant each $@_i$ has a left adjoint. Of course, this is because each $@_i$ is representable, as you observed.
1d	comment	Questions about epimorphisms and projectives in functor categories Each representable presheaf $\mathcal{I}(i, -)$ is free because it occurs as the image of $1$ under the left adjoint of the evaluation-at-$i$ functor. (This, in some sense, is the content of the Yoneda lemma.) That they are projective is easily shown to be a consequence of this, exactly like how free modules are projective. For the canonical projective covering of a presheaf $P$, just take the collection of all morphisms from any representable presheaf to $P$, and then take the amalgamation of all those.
1d	comment	a group is not the union of two proper subgroups - how to internalize this into other categories? My previous comments about toposes with enough points were incorrect; please disregard.
1d	comment	uniformization theorem - squares and circles The closed square is not a Riemann surface – it isn't even a manifold without boundary!
2d	comment	Grothendieck topology on pre/sheaves That defines, at best, a pretopology. However the sieve generated by any such family is a covering sieve in the canonical topology.
2d	comment	What are some examples of subtle logical pitfalls? Actually, the construction of JDH indicated here shows that Fact I is enough to construct, within any given model of ZFC, a transitive set that externally is a model of ZFC!
2d	comment	a group is not the union of two proper subgroups - how to internalize this into other categories? Actually, there's an easy counterexample in $\mathbf{Grp}$ as well, since internal groups there are just abelian groups.
May 22	comment	When is the pullback functor on sheaves faithful? It's true at the level of toposes and at the level of sheaves of abelian groups as soon as the map is surjective, but I think it is also true at level of quasicoherent sheaves if the morphism is faithfully flat. So what kind of sheaves are you asking about?
May 21	comment	A question regarding etale morphisms of affine varieties Well, the closed subvarieties are either the whole variety or finite sets of points, so those won't do either. And I don't think you really want to consider subvarieties that are neither open nor closed...
May 21	comment	A question regarding etale morphisms of affine varieties I suppose you mean to either have $p$ surjective, or "locally isomorphic" instead of "isomorphic".
May 20	comment	Does every category have a functor? Your answer is false if taken literally: $1$ does not embed into the empty category, and the category $1$ only has the identity endofunctor.
May 20	comment	Does every category have a functor? Of course, if $D$ is empty...
May 20	comment	Question about comultiplication This is formally dual to the following problem: Suppose $A$ is an algebra over the field $k$. Show that the multiplication map $\mu : A \otimes A \to A$ is an algebra map if and only if $A$ is commutative.
May 19	comment	ZF Extensionality axiom The converse is a logical tautology, however.
May 19	comment	Finishing the proof that $\textrm{Spec}K[X,Y]\setminus\{(X,Y)\}$ is not an affine scheme $f^*$ is an isomorphism, but whether or not it is the identity depends on the explicit construction of the structure sheaf etc.
May 19	comment	Finishing the proof that $\textrm{Spec}K[X,Y]\setminus\{(X,Y)\}$ is not an affine scheme No, you start with $f$ and then calculate $f^*$.
May 19	comment	Finishing the proof that $\textrm{Spec}K[X,Y]\setminus\{(X,Y)\}$ is not an affine scheme You can do that if you like, but that is not necessary.