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May
7
revised Connected spaces where all subsets are either open or closed
added 155 characters in body
May
7
revised Connected spaces where all subsets are either open or closed
edited body
May
7
asked Connected spaces where all subsets are either open or closed
May
6
comment A path to more advanced math topics.
I already upvoted it, but I’d like to verbally support the recommendation made by Hasan as well: If you know about matrices, determinants, vectors, vector products and such and you are intrigued by algebra, chances are you will find linear algebra extremely enlightening. It’s probably the most natural thing to go to if you want to get introduced to more abstract and structural mathematics. It is also helpful, if not necessary, for studying multivariate calculus and abstract analysis – which would be the next step on the calculus path.
May
6
comment Definition of a prime
“Now, $6$ is not a unit in $ℤ$” is what that should read.
May
6
comment formal power series expansion for square root
@OneWingedPterodactyl And the index in the sum is $n$, not $i$.
May
6
comment formal power series expansion for square root
Hm, your series suffers from index sickness.
May
6
revised Why does an isomorphism need to be a homomorphism?
added 2 characters in body
May
6
answered Why does an isomorphism need to be a homomorphism?
May
6
comment Why does an isomorphism need to be a homomorphism?
Also, I think this answer is missing the point of the question.
May
6
comment Why does an isomorphism need to be a homomorphism?
By saying “which admits a two-sided inverse” and leaving out the part where it’s mentioned that this inverse should be a homomorphism as well, you’re hiding the whole point you are making.
Apr
30
comment If $H$ is a cyclic group of even order, $H$ has exactly two elements which square to $1.$
There are finite abelian groups that only do have one element of order $2$, yet are not cyclic. Take $ℤ/2ℤ × (ℤ/3ℤ)^2$.
Apr
30
revised If $H$ is a cyclic group of even order, $H$ has exactly two elements which square to $1.$
added 398 characters in body
Apr
30
answered If $H$ is a cyclic group of even order, $H$ has exactly two elements which square to $1.$
Apr
30
comment If $ f : S^n \to R^n $ satisfies $f(-x)=-f(x)$ for all $x \in S^n $,then there exists $ y \in S^n $ with $f(y)=0 $
@SaikatBasu If $f$ isn’t assumed to be continuous, then the statement is obviously false. Just partition $S^n$ into antipodal sets $S^n = T \sqcup -T$ and take any function $f \colon T → ℝ^n$ with no zeros. Then extend it to $S^n$ such that $f(-x) = -f(x)$ $∀x ∈ T$.
Apr
28
comment Can all “standard” properties of the tensor product be proven from the universal property?
For example, if $ψ\colon V × W → V \otimes W$ is the bilinear map from the universal property, and $V$ and $W$ are free with bases $B ⊂ V$ and $C ⊂ W$, then for all vector spaces/moudles $Z$ and all maps $ψ(B × C) → Z$, there is a map $B × C → ψ(B×C) → Z$ and so there is exactly one bilinear map $V × W → Z$ extending this map and so there is, by the universal property of tensor products, exactly one linear map $V \otimes W → Z$ commuting with $ψ$ and the extended bilinear map on $V × W$, so extending the map $ψ(B × C) → Z$ - therefore, $ψ(B × C)$ is a basis of $V \otimes W$.
Apr
28
comment Can all “standard” properties of the tensor product be proven from the universal property?
I once thought about this during a long train ride. I’m pretty sure you can prove everything if you also invoke universal properties of free objects etc. But I certainly haven’t thought this through. I like this question and I hope it gets answered.
Apr
28
comment Prove that $V= \text{Im} f \oplus \ker{g}$
$\mathrm{img}~f ⊂ W$, so you should use an isomorphism sign in $V \cong \mathrm{img}~f \oplus \ker g$. Otherwise it gets confusing as there is such a thing as an internal direct sum which should be contrasted to the external direct sum.
Apr
27
revised Prove that if $A$ is both open and closed, $A=\mathbb R$.
edited tags
Apr
27
comment Prove that if $A$ is both open and closed, $A=\mathbb R$.
Definition of open/closed is crucial here.