Jul 29, 2014
2 notes
Let $$\mathcal C$$ be the category of semi-symplectic topological quantum paramonoids of Rice-Paddy type, satisfying the Mussolini-Rostropovich equations at infinity. Let $$X, Y$$ be objects of $$\mathcal C$$ such that the functors $$\text{Mor}_{\mathcal C}(X, \cdot)$$ and $$\text{Mor}_{\mathcal C}(Y, \cdot)$$ are isomorphic, as covariant functors from $$\mathcal C$$ to sets. Show that $$X$$ and $$Y$$ are isomorphic in $$\mathcal C$$.
god bless the sense of humor of whoever wrote that qual
Jul 27, 2014
0 notes

A quick lemma:

Lemma. Let $$f \in \mathrm{Hom}(A,B), g \in \mathrm{Hom}(B,C)$$ in an abelian category. Then $$-(g \circ f) = -g \circ f = g \circ -f$$ in $$\mathrm{Hom}(A,C)$$.

Proof. By the bilinearity of morphism-composition $$(- \circ -) = \circ(-,-)$$,

$- \circ (g,f) + (g,f) = 0,$

$\circ (-g,f) + \circ(g,f) = \circ(0,f) = 0,$

and

$\circ (g,f) + \circ (g, -f) = \circ (g,0) = 0.$

Combining the three equalities and dropping the $$\circ(g,f)$$’s gives the result. $$_\square$$

Golly gosh, I’m awful at these kinds of tricks.

Jul 26, 2014
1 note

On the existence of simple groups of arbitrary cardinality

What wonderful answers (and question by Pete Clark.)

Jul 16, 2014
0 notes

Some (easy) definitions:

Appropriated from Wiki:

An abelian category is a category that 1) enriched over Ab, so every hom-object is an abelian group and morphism-composition is bilinear, 2) admits a zero object and bi (direct) products, 3) admits kernels and cokernels, and 4) has the property that every monomorphism (resp. epi-) is the kernel (in the category-theoretic kernels-are-morphisms sense) (resp. cokernel) of some morphism.

(Examples: fix a ring $$R$$; then the category of 1) $$R$$-modules is an abelian category; 2) sheaves of abelian groups on a topological space; 3) abelian groups; 4) finite vector spaces over a field.)

A sequence of morphisms $$A_1 \overset{f_1}{\to} A_2 \overset{f_2}{\to} A_3 \cdots$$ is exact when $$\mathrm{im} f_i \simeq \ker f_{i + 1}$$ for each $$i$$.

A short exact sequence is an exact sequence $$A_1 \overset{f_1}{\to} A_2 \overset{f_2}{\to} A_3$$ where $$f_1$$ is a monomorphism (resp. $$f_2$$, epi); in an abelian category we write

$0 \to A_1 \overset{f_1}{\to} A_2 \overset{f_2}{\to} A_3 \to 0.$

Jun 13, 2014
536 notes

math textbooks are the best

Jun 7, 2014
527 notes
I have no idea why I’m using this letter, I can’t even pronounce it.
Mathematical Logic lecturer on the letter ‘xi’ (via mathprofessorquotes)
Jun 4, 2014
3 notes
For example, why is 6/2 = 3? We can take a 6-element set $$S$$ with a free action of the group $$G = \mathbb{Z}/(2)$$ and identify all the elements in each orbit to obtain a three-element set $$S/G$$.
John Baez, From Finite Sets to Feynman Diagrams
Jun 4, 2014
6 notes
Now, given a category C, we may ‘decategorify’ it by forgetting about the morphisms and pretending that isomorphic objects are equal. We are left with a set (or class) whose elements are isomorphism classes of objects of C. This process is dangerous, because it destroys useful information. It amounts to forgetting which road we took from x to y, and just remembering that we got there. Sometimes this is actually useful, but most of the time people do it unconsciously, out of mathematical naivete. We write equations, when we really should specify isomorphisms. ‘Categorification’ is the attempt to undo this mistake. Like any attempt to restore lost information, it not a completely systematic process. Its importance is that it brings to light previously hidden mathematical structures, and clarifies things that would otherwise remain mysterious. It seems strange and complicated at first, but ultimately the goal is to make things simpler.
John Baez, From Finite Sets to Feynman Diagrams
Jun 4, 2014
3 notes

There exists a surjection $$2^{N_0} \rightarrow \aleph_1$$

(appropriated from this discussion at math.SE)

Let $$f:\mathbb{N} \rightarrow \mathbb{Q}$$ be an enumeration of the rationals; since $$\forall \alpha <_{ON} \omega_1 \exists I \subseteq \mathbb{Q}$$ such that $$I \simeq \alpha$$ obtain an embedding $$\iota:\omega_1 \rightarrow 2^{\omega}$$ by $$\alpha \mapsto f^{-1}(I)$$; inverting this embedding and sending everything not in $$\mathrm{range}(\iota)$$ to, say, $$0$$, gives the required surjection.

Jun 4, 2014
2 notes

Every countable ordinal embeds into the rationals

Define by induction the refinement $$r(\alpha)$$ of an ordinal $$\alpha$$: set

$r_0(\alpha) = \alpha$ and $$r_n(\alpha) = r_{n-1}(\alpha)$$ with new elements inserted between every pair of elements $$a_1,a_2$$ such that $$\not \exists a_3 : a_1 < a_3 < a_2$$; refine the well-ordering of $$r_{n-1}(\alpha)$$ to reflect this. Then we have $r(\alpha) = \bigcup_{n < \omega} r_n(\alpha)$ is a countable dense linear order hence isomorphic to $$\mathbb{Q}$$, and $$r_0(\alpha) \hookrightarrow r(\alpha)$$ in the obvious way.

To obtain those new elements, take ordered pairs, i.e $$a_3 = \langle a_1,a_2 \rangle$$; since we only iterate up to $$\omega$$ this is well-defined and we’re fine.

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