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8h
comment how would you define the term “elementary” in the context of categories and sets?
Elementary here is meant in the sense of logic, i.e. something related finitary first order logic (as opposed to higher order logic and infinitary logic).
14h
awarded  Nice Question
16h
awarded  Popular Question
1d
comment For an inductive limit $X = \bigcup X_n$ of vector spaces, show that $X$ is complete if $X_n$ is complete for all $n$
Oh, I suppose I was thinking about Banach spaces rather than topological vector spaces in general. Still, the direct limit $\varinjlim_n \mathbb{R}^n$ has the topology induced by the obvious metric (because the inclusions are all isometries), and that is not complete.
1d
comment The semidirect product as a deformation of the direct product
Well, to be more precise: the semidirect product $G \rtimes_\phi H$ is the product $G \times H$ "deformed" by the parameter $\phi : H \to \mathrm{Aut}(G)$, and if you set $\phi$ to be the constant function with value $\mathrm{id}$, then you get back the ordinary direct product. How is this not already analogous to deformation quantisation?
1d
comment The semidirect product as a deformation of the direct product
What's wrong with the simple fact that the direct product is a special case of the semidirect product?
1d
comment Generalizing a statement about direct limits in the category of $A$-modules to other categories
The embedding is guaranteed to be fully faithful and preserve finite limits and finite colimits (in particular, kernels, cokernels, and finite direct sums), which is enough to establish many of the basic lemmas of homological algebra.
1d
comment Self-duality in a lattice
I think there is a counterexample where $X$ is a (finite) poset rather than a lattice.
1d
comment For an inductive limit $X = \bigcup X_n$ of vector spaces, show that $X$ is complete if $X_n$ is complete for all $n$
The claim seems to be false. Take, for instance, $X_n = \mathbb{R}^n$. Then the inductive limit is a real vector space of countably infinite dimension, and these are never complete (regardless of the topology).
1d
comment Generalizing a statement about direct limits in the category of $A$-modules to other categories
@dorebell No. The embedding does not preserve direct limits in general, so you can't use the embedding theorem to say anything about direct limits.
1d
awarded  algebraic-topology
2d
comment Are multi-valued functions a rigorous concept or simply a conversational shorthand?
One possible definition is given here, where I discuss them in connection to branch points.
2d
answered Is there a discrete initial topology on the set of real numbers?
2d
comment Is there a discrete initial topology on the set of real numbers?
Show that if $X$ has the initial topology with respect to a map $f : X \to Y$ and $Y$ has a second-countable topology, then so does $X$.
2d
comment Tensor product of arbitrary categories
Your characterisation of tensor products of vector spaces is somewhat convoluted. If you do it the normal way, you will see that the tensor product of categories is precisely the cartesian product. (For an easier version of this, try to work out what the tensor product of partially ordered sets is.)
2d
comment Is there a discrete initial topology on the set of real numbers?
That's not what I said. (The image of, say, the unique function $\mathbb{R} \to \{ 0 \}$ is certainly second countable, whether or not $\mathbb{R}$ is topologised discretely.) You need to use an explicit description of the initial topology.
2d
comment When do evaluation and the integral sign “commute”?
The subtlety is whether $g$ itself is a continuous, differentiable, smooth etc. function.
2d
comment How to prove that : $ \mathrm{Hom} ( A(G), H) \simeq \mathrm{Hom} (G , I(H)) $?
Yes, if $G \to A$ is a group homomorphism and $A$ is abelian, then $[G, G]$ is in the kernel. Next, can you show that $G / [G, G]$ is abelian?
2d
comment How to prove that : $ \mathrm{Hom} ( A(G), H) \simeq \mathrm{Hom} (G , I(H)) $?
@Bryan261 Based on your previous questions, it seems as if your background in mathematics has some gaps. I strongly urge you to say what you understand and what you don't – and "I don't understand anything" is not acceptable.
2d
comment Is there a discrete initial topology on the set of real numbers?
There is no such function. Observe that $\mathbb{R}$ with the usual topology has a countable basis, but $\mathbb{R}$ with the discrete topology does not.