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.
Best Answer
I'm adding an answer to this question because I both agree and disagree with trujello's answer in the comments on the question.
I agree that the definition Riehl gives is somewhat opaque for a beginner, but I definitely disagree that the definition is bad. This is the definition that I use intuitively now, though it certainly wasn't the one I first learned (partly because I never saw a clear definition when I was learning, only a lot of examples).
I also think that if you start off with learning this definition you'll avoid a couple common misconceptions that I encountered in the past. For example, this definition makes the role of the universal element clear. A product is not just an object, but an object with a specific family of projections satisfying a property.
Some philosophy on definitions
What makes a definition good? One possible viewpoint is that a good mathematical definition is one that abstracts a bunch of common situations in some area of mathematics and identifies the key feature that allows you to understand them as being all examples of the same underlying concept.
A result of this viewpoint is that the definition can seem opaque when you aren't so familiar with the situations being abstracted to see how they all fit into this same concept. (For example, I've recently been introduced to the concept of triangulated categories, whose examples are the stable homotopy category, derived categories of complexes, and modular representation theory, none of which are particularly familiar to me, which makes understanding the definition a bit more difficult.)
How then can we understand such definitions?
The answer I'll give is in two parts, and I'll try to provide these in this answer. First, understand the examples motivating the definition. A good algebra textbook doesn't just drop the definition of a group on students with no examples, and this definition of universal property is the same. It needs to be packaged with a long list of examples. Second, the connection between the motivating examples and the definition needs to be made explicit.
With that said, let's get to answering the real question asked above, how to understand Riehl's definition, and what exactly is going on with universal elements/their uniqueness.
Motivating examples of universal properties
A note that all of these examples all are described as having universal properties whether presented in Riehl's form or not. Riehl's definition uses the name universal property because that is the name already in use for all the examples. I will therefore sort of have to use it in both the more traditional sense and in Riehl's sense, although they do turn out to be equivalent.
Also, I have a slight quibble of the precise wording of her definition, which may or may not be contributing to your confusion about the uniqueness of the universal element. I would say that the representing object $c$ is a universal object, the element $x\in F(c)$ is the universal element, and the pair $(c,x)$ has the universal property that it represents the functor $F$.
Products (more generally limits)
Let $C$ be a category, $x,y\in C$ objects. A product of $x$ and $y$ is an object $x\times y$, and a pair of morphisms $\pi_1 : x\times y \to x$, $\pi_2:x\times y \to y$ such that the following universal property (not obviously in Riehl's sense) is satisfied:
For all objects $z$, with a pair of morphisms $p_1 : z\to x$ and $p_2:z\to y$, there is a unique morphism $(p_1,p_2): z\to x\times y$ such that $\pi_1(p_1,p_2) = p_1$, and $\pi_2(p_1,p_2)=p_2$.
Relating to Riehl's definition: We need a functor and a universal element. The functor here is $C(-,x)\times C(-,y)$, where $C(x,y)$ is notation for the morphisms from $x$ to $y$ in the category $C$. In other words, this functor sends an object $z$ to pairs of morphisms $(p_1:z\to x, p_2:z\to y)$.
A representing object for $C(-,x)\times C(-,y)$ consists of a pair of an object $x\times y$ and a universal element $$(\pi_1,\pi_2)\in C(x\times y,x)\times C(x\times y,y),$$ such that $(\pi_1,\pi_2)$ defines a natural isomorphism $$C(-,x\times y)\simeq C(-,x)\times C(_,y).$$ This property of the universal element translates into the fact that morphisms $f:z\to x\times y$ correspond bijectively to pairs of maps $(f_1:z\to x,f_2:z\to y)$ via $f\mapsto (\pi_1\circ f,\pi_2\circ f)$, which is the first version of the universal property given in the definition of the product.
Free abelian groups (more generally free objects)
Let $S$ be a set. Let $U$ be the forgetful functor from abelian groups to sets. Let $\newcommand\Z{\mathbb{Z}} \Z\{S\}$ denote the free abelian group on the set $S$, with $[s]$ denoting the basis element corresponding to a given element $s\in S$. The universal property (again, not obviously Riehl's definition, though equivalent) of $\Z\{S\}$ is that for any abelian group $A$, and any map $\phi:S\to UA$, there is a unique morphism $\tilde{\phi}:\Z\{S\}\to A$ such that $\tilde{\phi}([s])=\phi(s)$.
Relating to Riehl's definition The functor here is $\newcommand\Set{\mathbf{Set}} \Set(S,U-)$. The universal object should be $\Z\{S\}$, and the universal element should be a function $\psi : S\to U\Z\{S\}$, which is $\psi(s)=[s]$. The natural transformation corresponding to $\psi$ sends a morphism $\tilde{\phi}:\Z\{S\}\to A$ to the function $s\mapsto \tilde{\phi}([s])$. The fact that this natural transformation is a natural isomorphism yields the version of the universal property above.
Tensor products
Again, we'll give the traditional definition of the tensor product. The tensor product $V\otimes_k W$ of vector spaces $V$ and $W$ over a field $k$ is a vector space with a $k$-bilinear map $\otimes : V\times W\to V\otimes_k W$ such that for any $k$-bilinear map $\phi : V\times W\to U$, there is a unique linear map $\tilde{\phi}:V\otimes_k W\to U$ with $\phi =\tilde{\phi}\circ \otimes$.
Here the functor is $\newcommand\Bilin{\operatorname{Bilin}}\Bilin_k(V,W;-)$, the universal object is $\otimes$, the universal element is $\otimes$, and we see that the universal property is about a bijection between maps $\newcommand\Vect{\mathbf{Vect}}\Vect_k(V\otimes_kW,U)$ and $\Bilin_k(V,W;U)$.
There are of course many more examples of universal properties, but I'll leave this here.
Universal elements and their uniqueness
Firstly, universal objects and universal elements are often not unique, but they are unique up to unique isomorphism in a sense that I'll explain below.
First, an example of nonuniqueness. In the tensor product example we can take any automorphism of $V\otimes_k W$ as a $k$-vector space, $\alpha$, and $\alpha\circ \otimes$ will also be a universal element. The point to take away from this is that the tensor product is not just the object $V\otimes_k W$ (despite the somewhat misleading notation), it is the pair $(V\otimes_k W,\otimes)$. I.e. the pair of a universal object and a universal element. These pairs are not unique, but they are unique up to unique isomorphism, which is just about the best you can get in category theory.
Suppose $(c,x)$ is a representing pair for a functor $F$, and suppose $(c',x')$ is also a representing pair for $F$. Then there is a unique isomorphism $c\to c'$ such that $F(c\to c')$ sends $x$ to $x'$. The proof of this is an application of the Yoneda lemma: We have natural isomorphisms $$C(c,-)\simeq F\simeq C(c',-),$$ where the following elements correspond $$1_c\leftrightarrow x\leftrightarrow \phi:c'\to c$$ $$\psi:c\to c'\leftrightarrow x'\leftrightarrow 1_{c'}.$$ Applying $\psi$ to the first correspondence, and $\phi$ to the second correspondence, we get $$\psi \leftrightarrow F(\psi)(x)\leftrightarrow \psi\phi$$ $$\phi\psi\leftrightarrow F(\phi)(x')\leftrightarrow \phi.$$ Comparing these to the first pair of correspondences, we see that $\psi\phi = 1_{c'}$, $\phi\psi = 1_c$, and $F(\psi)(x) = x'$, which proves existence of the claimed isomorphism. Uniqueness is because if $F(\beta)(x)=x'$, then we would get $\beta \leftrightarrow x' \leftrightarrow \beta\phi$, which forces $\beta = \psi$.