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Sorry for the confusing title. I just started learning category theory and am very confused about the concept "universal property". I am not even sure whether my "proof" is a proof or is just a restating of the original problem.

Here is a problem from Vakil's notes of Algebraic Geometry.

Exercise 1.3.D. Verify that $A\rightarrow S^{-1}A$ satisfies the following universal property: $S^{-1}A$ is initial among $A$-algebras $B$ where every element of $S$ is sent to an invertible element in $B$.

My attempt: I searched and found this question on MSE: https://math.stackexchange.com/questions/1816224/localization-and-the-universal-property?rq=1.

So I constructed my "proof" based on that:

We consider a category $\mathscr{C}$ with objects as pairs of, e.g., $(S^{-1}A, A\xrightarrow{i_s} S^{-1}A), (B, A\xrightarrow{\pi_B} B), (C, A\xrightarrow{\pi_C} C)$. The objects are pairs whose first components are $A$-algebras, and second components are ring maps from $A$ to the corresponding $A$-algebras, in which elements from $S$ are sent to an invertible element. Notice that $S$ is in $A$. For any two objects $B, C$, the morphisms are ring maps between $B$ and $C$. We have the following diagram:

enter image description here

We see that any map $\pi: A\rightarrow B$ where every element of $S$ is sent to an invertible element must factor uniquely through $i_s: A\rightarrow S^{-1}A$.

My question: Could anyone verify my "proof"? I am not sure how this universal property could be used.

Edit:

According to @Hoot's suggestion, I will define $\phi$ and prove that it is unique. Define $\phi: S^{-1}A\rightarrow B$ by $\phi(a/s)=\pi(a)\pi(s)^{-1}$, for $s\in S, a\in A$.

We prove that $\phi$ is well-defined. Let $a_1/s_1=a_2/s_2$, i.e., $s(a_1s_2-a_2s_1)=0$ for some $s\in S$. This implies $\pi(a_1s_2-a_2s_1)=0$. We have $\phi(a_1/s_1)=\pi(a_1)\pi(s_1)^{-1}, \phi(a_2/s_2)=\pi(a_2)\pi(s_2)^{-1}$. Apparently they are equal. By some similar computations we can prove it is a ring homomorphism.

It remains to show that $\phi$ is unique. Let $\psi: S^{-1}A\rightarrow B$ be another ring homomorphism, such that $\pi=\psi\circ i_s$. Then $\pi(a)=\psi(a)$ for all $a\in A$. Since it is homomorphism, $\psi(a/s)=\psi(a)\psi(s)^{-1}=\pi(a)\pi(s)^{-1}$. This shows that $\psi=\phi$.

Another question:

In the notes the author says every map $A\rightarrow B$ that sends elements in $S$ to an invertible element factors uniquely through $A\rightarrow S^{-1}A$. I'm still confused about this sentence. Shouldn't we say it factors uniquely through $S^{-1}A\rightarrow B$, since that is the unique universal map?

Thank you for your help!

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    $\begingroup$ I think your "proof" is simply asserting the universal property. $\endgroup$ – Jakob Hansen Jun 25 '16 at 13:38
  • $\begingroup$ @JakobHansen: Thanks, that's what I suspected:(. So how to really prove it? $\endgroup$ – KittyL Jun 25 '16 at 16:06
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    $\begingroup$ Say you're in the situation of the diagram. You need to show that $\phi$ exists and that no other $\phi$ works. Can you start by trying to define a $\phi$? The first issue is whether this thing is well defined. $\endgroup$ – Hoot Jun 25 '16 at 16:59
  • $\begingroup$ @Hoot: Thank you! Now I know what to try. $\endgroup$ – KittyL Jun 25 '16 at 17:17
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    $\begingroup$ I got confused, sorry. Your organization was not the usual one and I went into autopilot. You're totally correct and it follows because $\pi(s)$ is a unit; I would make that explicit in the proof. One should check that $\phi$ is a homomorphism but you probably don't want to do that here. Also it's quite easy. The uniqueness looks good too. $\endgroup$ – Hoot Jun 25 '16 at 21:34
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Setup

Following page 32 of Vakil's notes,let S be a multiplicative subring of a ring A; i.e., $1 ∈ S ∧ x,y ∈ S ⇒ x · y ∈ S$. Then we consider “formal fractions”,

S⁻¹A ≔ { a / s ∣ a ∈ A , s ∈ S }

The property we're interested in is

𝒫 : A-algebra → Bool
𝒫 f ≔ for every e in S, f e ∈ B is invertible 

Want to show: S⁻¹A is initial among A-algebras B where every element of S is sent to an invertible element in B

So we're interested in the category whose objects are pairs (B, f) where B is a ring and f : A → B is a ring morphism satisfying 𝒫.

Note that $S⁻¹A$ is such an algebra due to the emebeding $i : a ↦ a / 1$ since we know $1 ∈ S$. Moreover we have that $𝒫 i$ holds since $(s / 1)⁻¹ = 1 / s$ in $S⁻¹A$ for $s \in S$ : it is clear that $1/1$ is the unit of multiplication and so we need to show $(s / 1) · (1 / s) = 1 / 1$. Indeed,

   (s / 1) · (1 / s) = 1 / 1
≡{ multiplication in S⁻¹A }
   s · 1 / 1 · s = 1 / 1
≡{ ring arithmetic }
   s / s = 1 / 1
≡{ equality in S⁻¹A }
   ∃ s' : S • s' · (1 · s - s · 1) = 0
≡{ ring arithmetic }
   ∃ s' : S • s' · 0 = 0
≡{ ring arithmetic }
   true

Now the natural definition of an arrow between two such objects is a ring morphism of the underlying rings that makes the obvious diagram commute: f : (B, g) ⟶ (C, h) iff f : B → C is a ring map with f ∘ g = h.

  A --g--->B
  |       /
  |      /
  h     / f
  |    /
  |   /
  v  /
  C</

Defining the needed morphism

Given an A-algebra $f : A → B$ satisfying $𝒫$, let us show that there is an arrow $(S⁻¹A, i) ⟶ (B, f)$.

Given an element $a / s : S⁻¹A$, we know that $f \ s$ is invertible by assumption $𝒫 \ f$ and so we can form the expression $f a · (f s)⁻¹ : B$. Let $⟨ f ⟩$ be the name of this operation; i.e., $⟨ f ⟩ : S⁻¹A → B$ with $⟨ f ⟩ (a / s) = f a · (f s)⁻¹$. Since / is just formal notation for pairs, in this context, that gets reified into an actual division by ⟨_⟩, let us write ÷ for a legitimate division in a ring: $x ÷ y = x · y ⁻¹$. Then, $⟨ f ⟩ (a / s) = f a ÷ f s$. Notice that we do not know if $s⁻¹$ exists and so cannot use ring morphism properties to rewrite this last expression as $f (a / s)$.

Anyhow, for ⟨ f ⟩ to be an arrow in our category we must show that is

  • satisfies 𝒫
  • is a ring morphism
  • satisfies the triangle diagram
  • and is unique, so that we have intiality.

Satisfies 𝒫

Let $a, s$ be arbitrary, then

⟨ f ⟩ (a / s) ⁻¹ = ⟨ f ⟩ (s / a)

Indeed,

   ⟨ f ⟩ (a / s) · ⟨ f ⟩ (s / a)
={ definition of angels }
  (f a ÷ f s) · (f s ÷ f a)
={ division notation }
  f a · (f s)⁻¹ · f s · (f a)⁻¹
={ ring arithmetic }
  f a · 1 · (f a)⁻¹
={ ring arithmetic }
  1

Ring morphism

For example, for additivity:

  ⟨ f ⟩ (a/s + b/t)
={ definition of + in S⁻¹A }
  ⟨ f ⟩ ( (t · a + s · b)/(s · t) )
={ definition of angels }
  f (t · a + s · b) ÷ f (s · t)
={ f is a ring morphism }
  (f t · f a + f s · f b) ÷ (f s · f t)
={ ring arithmetic }
  (f t · f a) ÷ (f s · f t)
  + (f s · f b) ÷ (f s · f t)
={ ring arithmetic:

     (f t · f a) ÷ (f s · f t)
   ={ definition of ÷ }
     f t · f a · (f s · f t)⁻¹
   ={ ring arithmetic }
     f t · f a · f t ⁻¹ · f s ⁻¹
   ={ f ring homomorphism }
     f (t · a) · f t ⁻¹ · f s ⁻¹
   ={ definition }
     ⟨ f ⟩  ( t · a / t) · f s ⁻¹
   ={ lemma; see below; and ring arithmetic using f ring morphism }
     f a · f s ⁻¹
   ={ notation }
     f a ÷ f s

   lemma:
     t · a / t = a / 1
   ≡{ equality in S⁻¹A }
     ∃ s : S • s · (1 · t · a - t · a) = 0
   ≡{ ring arithemtic }
     ∃ s : S • s · 0 = 0
   ≡{ ring arithmetic }
     true

   Similarly for the right argument to +.

   Anyhow, back to the main calculation.
}
  f a ÷ f s + f b ÷ f t
={ definition }
  ⟨ f ⟩ (a / s) + ⟨ f ⟩ (b / t)

I will not check the other properties ─I'm not as vested in this problem as you!

Arrow

It is also an arrow in the category; indeed: for arbitrary a,

   ⟨ f ⟩  (i a)
={ definition of i }
   ⟨ f ⟩ (a / 1)   
={ definition of ⟨⟩ }
   f a ÷ f 1
={ f ring morphism }
   f a ÷ 1
={ ring arithmetic }   
   f a

Whence $⟨ f ⟩ ∘ i = f$

Uniqueness

It remains to show that $⟨ f ⟩$ is unique with these properties.

That is, given any ring morphism $G : S⁻¹A → B$ with $G ∘ i = f$, let us show that it equals $⟨ f ⟩$.

  ⟨ f ⟩ (a / s)
={ definition }  
   f a ÷ f s
={ assumption G ∘ i = f }
   G (i a) ÷ G (i s)
={ definition of i}
   G (a / 1) ÷ G (s / 1)
={ notation }
   G (a / 1) · G (s / 1) ⁻¹
={ G ring morphism }
   G ( (a / 1) · (s / 1)⁻¹ )
={ first lemma about inverses of formal fractions }
 G ( (a / 1) · (1 / s) )
={ multiplication in S⁻¹A }
  G (a · 1 / 1 · s)
={ ring arithmetic }
  G (a / s)

Hence, by extensionality, G = ⟨ f ⟩ sweetums :-)

Conclusion

At the end of the problem statement they give rephrasings of the desideratum, which we'll restate for fun.

Using the above formalisation and usual category theory definition of initiality, we have

  S⁻¹A is initial among A-algebras sending every element of S to an invertible element
≡
  ∀ B ring, f : A → B ring-map satisfying 𝒫 • ∃₁ f' : S⁻¹A → B • f' ∘ i = f ∧ 𝒫 f'

But by construction this f' is obtained by the angle brackets; we're doing a process known as skolemisation. So this equivales

∀ B ring, ring maps f : A → B and g : S⁻¹A → B •
  g = ⟨ f ⟩ ≡ 𝒫 f ∧ g ∘ i = f

This' loosey-goosey true since: for ⇒ to hold we need ⟨ f ⟩ to be defined which means we need 𝒫 f to hold and we need to know that g is an arrow in the category which means g ∘ i = f; conversely, we know that g ∘ i = f has the unique solution ⟨ f ⟩ which is defined since we have 𝒫 f. A formal proof is not difficult either:

 𝒫 f ∧ g ∘ i = f
⇒{ “Defining the needed morphism” and “uniqueness” } 
  g = ⟨ f ⟩
⇒{ composition and “arrow” }
  g ∘ i = f
≡{ logic }
  g ∘ i = f ∧ g ∘ i = f
⇒{
    claim: g ∘ i = f ⇒ 𝒫 f
    Proof: let s : S be arbitrary, then
      we show (f s)⁻¹ = g (1 / s).

        f s · g (1 / s)
      ={ assumption and definition of i }
        g (s / 1) · g (1 / s)
      ={ g ring morphism }
        g ( s/1 · 1/s )
      ={ very first calculation in “setup” }
        g (1 / 1)
      ={ g ring morphism }
        1
  }  
  𝒫 f ∧ g ∘ i = f

Anyhow,

∀ B ring, ring maps f : A → B and g : S⁻¹A → B •
  g = ⟨ f ⟩ ≡ 𝒫 f ∧ g ∘ i = f

Read ⇒: a ring map S⁻¹A → B is the “same thing as” (ie corresponds to) a ring map A → B satisfying 𝒫

Read ⇐: any ring map A → B sending elements of S to invertible elements factors through i : A → S⁻¹A and does so uniquely.

Intiality really just says that $i$ is the best map satisfying 𝒫 and any other such map must be “further away” (in that it factors through $i$).

Hope this helps! :-)

PS. Notice that since $g ∘ i = f ⇒ 𝒫 f$, we could simplify the equivalence to just $g = ⟨ f ⟩ ≡ g ∘ i = f$, but then those English renditions are no longer immediate.

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    $\begingroup$ @Thank you! That is very clear and very helpful! $\endgroup$ – KittyL Jun 26 '16 at 10:03

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