Reputation
15,674
Next privilege 20,000 Rep.
Access 'trusted user' tools
Badges
1 13 42
Impact
~141k people reached

11h
revised Left and Right Inverses with semigroups
added 615 characters in body
11h
comment Left and Right Inverses with semigroups
Well, this isn't a proof, it's just supposed to help you produce one. Since you're new to this, let me give an example of how a writeup might look.
11h
comment Categories for the working mathematician exercises III 1
Just to comment that the "forgetting addition" functor maps $R$ to its group of units, so is rather different from the functor here.
11h
answered Categories for the working mathematician exercises III 1
12h
answered Left and Right Inverses with semigroups
13h
comment Is there an analogue of Eilenberg-Maclane spaces for homology?
For ordinary homology $E$ is just the Eilenberg-Maclane spectrum again, equivalently $H_n(X;\pi)=\varinjlim\pi_{n+k}(K(\pi,k)\wedge X_+)$
13h
comment What are higher derivatives?
I still think it's reasonable to say that the Hessian is not a tensor when it doesn't transform by a pointwise formula under coordinate changes. It's just that in the multivariable calculus case you have a preferred choice of coordinates, so these problems are not as acute.
13h
comment What are higher derivatives?
It seems to me that Wikipedia's comment comes from manifold theory: higher derivatives don't transform well under change of coordinates.
1d
comment Does the product functor preserve quotient maps?
The situation you wonder about in 2 is very common in cartesian closed categories, where every product functor is a left adjoint, at least if we let "quotient" mean "regular epimorphism", as it does here.
1d
comment Does the product functor preserve quotient maps?
The second paragraph seems more complicated than necessary. Locally compact spaces are exponentiable, so the corresponding product functor is a left adjoint, and thus preserves colimits, such as quotients.
May
27
comment Real Manifold … Complex Coordinates?
You wrote $x^4\sin t$ for $x^4\cos t$ in your real coordinate expression.
May
27
answered Tensor product of homology equivalences
May
27
comment Unifying Connection Between Topological Embeddings and Quotient Maps
@goblin I guess there's another question here, which could be interpreted in this context as "why does the same property turn both monos and epis extremal?" I think the reason is simply that the notion "continuous and open" is dual to the notion "open and continuous," but I'd also be interested to see another take.
May
26
comment Unifying Connection Between Topological Embeddings and Quotient Maps
@goblin I took the question to be "why are quotients and embeddings so closely related," and the OP apparently disagrees with you about whether this answers the OP's question.
May
26
revised Unifying Connection Between Topological Embeddings and Quotient Maps
edited tags
May
26
answered Unifying Connection Between Topological Embeddings and Quotient Maps
May
24
comment $\mathbb{Z}G\otimes_{\mathbb{Z}N} \mathcal{N}= \mathcal{N}^G$ is a non-trivial idempotent ideal in $\mathbb{Z}G$
This doesn't seem to be true in case $G=\mathbb{Z}/2$ and $N=G$. Then if $\mathbb{Z}/2=\langle \eta\rangle$, an element of $\mathcal{N}\cong \mathbb{Z}G\otimes_{\mathbb{Z} G} \mathcal{N}$ is of the form $a-a\eta$, and products of such elements have both coefficients even.
May
22
answered Is it possible to develop differential geometry without points?
May
20
comment Constructing the natural numbers without set theory.
Perhaps nitpicky (though the downvote is not from me) but Peano gave axioms for the natural numbers, not a construction of a model of those axioms.
May
19
awarded  soft-question