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Jun
2
comment Can a free group over a set be constructed this way (without equivalence classes of words)?
If you construct the free group using the left adjoint of the forgetful functor $\mathbf{Grp} \to \mathbf{Set}$, you will also not need to mention any equivalence classes of words or reduced words...
Jun
1
answered Is there a more elementary proof of this special case of Riemann-Roch?
Jun
1
comment Is there a more elementary proof of this special case of Riemann-Roch?
Of course, it remains to be shown that elliptic curves are not rational curves. This is usually done using a genus argument, but perhaps the OP does not want to use that...
Jun
1
comment Why are complete lattice homomorphisms defined that way?
Yes. But I'm sure you could have worked that out on your own.
Jun
1
answered A domain with only a (non-zero) prime ideal
May
31
revised Morphism of finite type between affine schemes is quasi-projective
edited tags
May
31
comment Morphism of finite type between affine schemes is quasi-projective
That doesn't look right. Why should $B$ be a graded ring?
May
31
comment The image of a morphism between affine algebraic varieties.
No. Let $V = \mathbb{A}^1$ and let $W$ be a point.
May
31
comment Why are complete lattice homomorphisms defined that way?
The same way as in varieties of finitary algebras: they preserve the various operations of the algebra. (The operations of a complete lattice are, of course, all the $\kappa$-ary joins and meets, as $\kappa$ varies over all cardinals.)
May
31
comment Why are complete lattice homomorphisms defined that way?
The standard definition is what you would get if you regard the theory of complete lattices as a variety of infinitary algebras. Is that not a good enough reason?
May
30
comment When terminal objects are separators?
Your claim is false even in $\mathbf{Set}$: take $A = \emptyset$. I doubt there is any good characterisation of the terminal object being a separator.
May
30
comment Intersection of Closed and Compact Set is Closed
@user45099 You have misunderstood Matt E's post. If $K$ is any subset whatsoever and $L$ is closed, then $K \cap L$ is closed as a subset of $K$.
May
30
comment The “set” of equivalence classes of things.
You have to be careful with what you mean by ‘cardinality’. For instance, let $Y$ be a set of disjoint pairs that has no choice function, let $X$ be the union of the pairs in $Y$, and let $X' = Y \times 2$. Then, assuming $Y$ can be well-ordered, then so can $X'$. But $X$ cannot, for otherwise we would have a choice function. In particular $X$ and $X'$ cannot be in bijection.
May
30
comment The “set” of equivalence classes of things.
The collection you describe is not literally a set, however. The standard way to avoid the axiom of choice for these things is to use Scott's trick.
May
30
comment some question of generic point
The second definition is not quite correct: it should also say that every point is a specialisation of $x$.
May
30
revised Subobjects and quotient objects in category of topological spaces
added 127 characters in body
May
29
answered Subobjects and quotient objects in category of topological spaces
May
29
comment If fundamental group of a compact path connected space is finite then its universal cover is compact
True. Local homeomorphism + finite fibres + surjective + connected base.
May
29
answered If fundamental group of a compact path connected space is finite then its universal cover is compact
May
29
comment If fundamental group of a compact path connected space is finite then its universal cover is compact
Sorry, that was a thinko. I confused it with the fact that local homeomorphisms (over a connected base) with finite fibres are covering maps.