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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.
Apr
27
answered Ring Homomorphism Defintion
Apr
26
asked What are the fixed points of $f_ c = c · \sin$ for $c > 1$?
Apr
26
comment If an isomorphism can be expressed as a composition of morphisms, what can we say about its components?
Think of the counterexample in $\mathrm{Ab}$ as a counterexample in $\mathrm{Set}$.
Apr
26
comment If an isomorphism can be expressed as a composition of morphisms, what can we say about its components?
We cannot. Think of categories with a null object, like the category of abelian groups. (And it doesn’t work for most concrete categories like $\mathrm{Set}$ itself.)
Apr
26
revised Show that $B/Q$ is integral over $A/P$
edited tags
Apr
26
answered Show that $B/Q$ is integral over $A/P$
Apr
25
answered Remember the implicit function theorem
Apr
25
comment $\dim (W_{1} \cap W_{2}) = \dim W_{1}$ implies $W_{1} \subset W_{2}$?
I didn’t mean to question your competence on linear algebra. To me it seemed like the problem was meant to exercise how to handle bases and base lengths and reason with them. From this point of view, I merely found “no need to reason about bases” to be false. That the questioner accepted the hint-answer, on the other hand, speaks for your interpretation.
Apr
25
comment $\dim (W_{1} \cap W_{2}) = \dim W_{1}$ implies $W_{1} \subset W_{2}$?
I think there is very much need to reason about bases. The statement doesn’t hold if you replace vector spaces with free modules and dimensions with rank, or if you go to the infinite-dimensional. You strongly use the fact that bases of finitely generated vector spaces have a unique length and are characterized by being maximal linearly independent systems. The answer you have given just subsumes all that in the step “If in addition it has the same (finite) dimension as $W_1$, then it must be all of $W_1$”.
Apr
25
answered $\dim (W_{1} \cap W_{2}) = \dim W_{1}$ implies $W_{1} \subset W_{2}$?
Apr
25
comment $\dim (W_{1} \cap W_{2}) = \dim W_{1}$ implies $W_{1} \subset W_{2}$?
You mean $β ⊂ W_1$.