#abstract algebra

CAN YOU ALL PLEASE SUGGEST ME FROM WHICH ONE OF MY PAPERS SHOULD I CHOOSE A TOPIC FOR MY RESEARCH PROJECT????????????

Topology or Abstract Algebra or Real Analysis

On the Subtle Joy of Injectivity

Given a category C alongside a class J of morphisms in C we can consider the full subcategory Inj(J) of C constructed by restricting to J-injective objects, namely those objects A in C such that for every monomorphism p : B -> E in J and every morphism f : B -> A in C there exists a morphism f⁺ : E -> A in C such that f⁺ o p = f.

We can characterise a lot of interesting mathematical categories as Inj(J) for some appropriately chosen class J.

Example: Consider the category of directed graphs whose objects are pairs (V, E) where V is a set of vertices and E is a binary relation on V whose elements are called edges, and whose morphisms f : (V, E) -> (V', E') are functions f : V -> V' such that (x,y) ∈ E implies (f(x), f(y)) ∈ E'. Observe that a directed graph is reflexive and only if it is injective with respect to ( { 0 } , ∅) -> ( { 0 }, { (0 , 0) } ) It is symmetric if and only if it is injective with respect to ( { 0, 1 } , { (0, 1) } ) -> ( { 0 , 1 }, { (0 , 1), (1, 0) } ) It is transitive if and only if it is injective with respect to ( { 0, 1, 2 } , { (0, 1), (1, 2) } ) -> ( { 0 , 1 , 2 }, { (0 , 1), (1, 2), (0, 2) } )

Hence, we can then say that the category of sets with equivalence relations is just Inj( { these three morphisms } ). Similarly we could say that a ring is commutative if and only if it injective with respect to ℤ{ x, y } -> ℤ { x , y } / ( xy - yx ) where ℤ{...} here is a free non-commutative ring.

Obstructions to Being Injective

Observe that for every object X we have an inclusion Hom(E, X) o J ⊆ Hom(B, X). Injectivity ensures this is an equality. Let us take free abelian groups to have ℤ[ Hom(E, X) o J ] ⊴ ℤ[ Hom(B, X) ] Hence we can define the injectivity obstruction group as G(X) := ℤ[ Hom(B, X) ]/ℤ[ Hom(E, X) \circ J ] With the enjoyable property that G(X) ≅ 0 iff X is injective with respect to at least one morphism in J.

Connectivity

A pleasing feature of singular homology is that we can actually prove somethings about these singular homology groups. So let us try and prove something about our injectivity obstruction group.

Oh ye! So, you know how when you take any real number, and you square it, even if the real number was negative, you wind up with a nonnegative number? Like, if I want to square two (2^2), I get 2*2, which is 4. If I want to square negative two ((-2)^2), I get (-2)*(-2), which is again 4, because the negatives cancel each other out.

I can do this with any number! 0^2 is 0*0 is 0 (nonnegative). (-1)^2 is (-1)*(-1) is 1 (positive, and therefore nonnegative). pi^2 is just pi^2, because pi is transcendental (a special form of irrational that I could go on about, but that takes a bit of a tangent - but basically, it means that if I plug pi into any polynomial with rational coefficients, then there's no way to get 0).

So, we have a well-defined function that can take any real number, and map it to another real number: f(x) = x^2. That's squaring!

Now, what if we wanted to do the reverse?

Well, we have a function for that, too! The square root function: f(x) = sqrt(x).

This takes in a number x, and it outputs the "primary square root" of x. When x is a nonnegative real number, it outputs the nonnegative real number y such that y^2 = x. So, we have sqrt(0) = 0, because 0*0 = 0. We have sqrt(4) = 2, because 2*2 = 4.

We have sqrt(2) as its own irrational number (somewhere close to 1.41...), with sqrt(2) defined as the nonnegative number x such that x^2 = 2. Similarly, sqrt(5) is the nonnegative number x such that x^2 = 5.

This doesn't address what happens when we want to do sqrt(x) where x is negative. What if I want to do sqrt(-4)? In other words, how do I find an x such that x^2 = -4?

x can't be -2, because, as we said earlier, (-2)*(-2) = +4, not -4.

That's where the idea for i comes in! i is defined as the primary square root of -1: that is, i is defined such that i^2 = -1.

When we first encounter i in classes that talk about imaginary numbers, we normally see this written as i = sqrt(-1). But that leads to the question: which root are we picking? sqrt(-1)^2 = -1, but also, (-sqrt(-1))^2 = -1.

As it turns out, if we base our number system on -i rather than on i, we get basically the same number system. So, rather than force a choice for i, which could be sqrt(-1) or could equivalently be -sqrt(-1), we define i such that i^2 = -1.

This lets us play with imaginary numbers! sqrt(-4) is now 2i. sqrt(-pi) is now i*sqrt(pi).

We can even take the square root of imaginary numbers! That's just asking the question, what is x such that x^2 = i, for example?

As it turns out, we need a number with both a real part and an imaginary part to answer that question: in other words, we need a complex number! A number a + bi, where a and b are real numbers!

And once we have complex numbers, we essentially have everything we need to build functions made out of polynomials and roots, without having to restrict the input and output!

So! TLDR: the idea of i comes from the fact that we can't take the square root of a negative number and come out with a real number. So, we need i to be one of the solutions to the equation x^2 = -1. Rather than write "i = sqrt(-1)" (no good, forces us to choose between sqrt(-1) and -sqrt(-1), why are we pitting our children against each other, -sqrt(-1) deserves to shine), we write "i^2 = -1" (good, does not force us to choose between these siblings, we love our children equally and we show it by not forcing sqrt(-1) into the spotlight and -sqrt(-1) into the shadows, they can share i).