Alright. Willis Todd being an abusive father to Jason is a trope often utilized. Comparing this version of him to Bruce's reactions to Red Hood is fantastic. Lots to analyze there.
However, I raise you. There needs to be more fanwork addressing the classism behind Willis Todd being characterized as an abusive alcoholic. In some version of canon, Willis Todd was a good dad in a shitty situation. He was poor, his wife (Catherine) was sick, and he had a newborn baby he needed to provide for. In this horrid situation, where he has no family to fall back on and no higher education to obtain a decent well-paying job, he tries to get quick money. He's desperate to keep both his wife and son alive.
Catherine turns to drugs because it's easier and cheaper to buy drugs than healthcare. The pain she experiences is debilitating, and she'd do anything to not feel pain for one godsdamned second. Unfortunately, this turns into an addiction.
This ultimately shapes the way that Jason views crime. Bruce, while he may be sympathetic to individuals who resort to crime to pay their bills, will not understand huddling in Crime Alley in the dead of winter as he debates whether to buy food or pay for heating. He won't understand the bitterness, hatred, pain, and resignation of never having enough money to survive as you get chewed up again and again.
If Jason's dad is just an abusive criminal, that not only perpetuates the notion that all criminals are evil, but it will shape how Jason views those who commit crime. Breaking the law doesn't make someone bad. There's plenty of reasons people commit crime, whether to survive, protect someone, or something else. The issue, especially in Gotham, is the system that perpetuates wealth inequality through bribes and unethical governmental practices.
Anyway, I think Jason's Red Hood is more fleshed out if it accounts for him acknowledging the desperation behind goons and small-time criminals because he grew up without other options.
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it feels weird to frame Ashton's current drive as wanting to be The Most Special and Chosen with this implication that he's being selfish or entitled with what he wants. Ashton, who can count the really good days in their life on their hands, who's been in constant pain for god knows how long. Ashton, a blank slate, deprived even of the meaning that comes from a history and identity. Ashton, deemed so worthless that even a group that called themselves the Nobodies abandoned him.
Ashton was not full and asking to feast; he was starving and asking to be fed. They weren't begging the gods to make them a champion and hand them a spell list, they just wanted someone to pick them up when they fell down for the first time in their life. instead, the gods (in any shape or form he could recognize) left him alone until the moment he become a problem for them.
of course he resents them. and of course he's grasping on to the first answers he's found, the first indication that he's ever mattered. it's fascinating to watch Ashton try to feel empowered by their world-altering burden instead of overwhelmed, and make decisions they'll think were fucking stupid in 20 episodes, and just overall try and tie his disparate parts together into one functioning whole, even if he's not really succeeding. especially if he's not really succeeding! we're not watching Ashton fly too close to the sun, asking for too much and getting burnt. we're watching them trying, clumsily and haphazardly, to pick up their broken pieces and create themselves anew.
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Welcome to the premier of One-Picture-Proof!
This is either going to be the first installment of a long running series or something I will never do again. (We'll see, don't know yet.)
Like the name suggests each iteration will showcase a theorem with its proof, all in one picture. I will provide preliminaries and definitions, as well as some execises so you can test your understanding. (Answers will be provided below the break.)
The goal is to ease people with some basic knowledge in mathematics into set theory, and its categorical approach specifically. While many of the theorems in this series will apply to topos theory in general, our main interest will be the topos Set. I will assume you are aware of the notations of commutative diagrams and some terminology. You will find each post to be very information dense, don't feel discouraged if you need some time on each diagram. When you have internalized everything mentioned in this post you have completed weeks worth of study from a variety of undergrad and grad courses. Try to work through the proof arrow by arrow, try out specific examples and it will become clear in retrospect.
Please feel free to submit your solutions and ask questions, I will try to clear up missunderstandings and it will help me designing further illustrations. (Of course you can just cheat, but where's the fun in that. Noone's here to judge you!)
Preliminaries and Definitions:
B^A is the exponential object, which contains all morphisms A→B. I comes equipped with the morphism eval. : A×(B^A)→B which can be thought of as evaluating an input-morphism pair (a,f)↦f(a).
The natural isomorphism curry sends a morphism X×A→B to the morphism X→B^A that partially evaluates it. (1×A≃A)
φ is just some morphism A→B^A.
Δ is the diagonal, which maps a↦(a,a).
1 is the terminal object, you can think of it as a single-point set.
We will start out with some introductory theorem, which many of you may already be familiar with. Here it is again, so you don't have to scroll all the way up:
Exercises:
What is the statement of the theorem?
Work through the proof, follow the arrows in the diagram, understand how it is composed.
What is the more popular name for this technique?
What are some applications of it? Work through those corollaries in the diagram.
Can the theorem be modified for epimorphisms? Why or why not?
For the advanced: What is the precise requirement on the category, such that we can perform this proof?
For the advanced: Can you alter the proof to lessen this requirement?
Bonus question: Can you see the Sicko face? Can you unsee it now?
Expand to see the solutions:
Solutions:
This is Lawvere's Fixed-Point Theorem. It states that, if there is a point-surjective morphism φ:A→B^A, then every endomorphism on B has a fixed point.
Good job! Nothing else to say here.
This is most commonly known as diagonalization, though many corollaries carry their own name. Usually it is stated in its contraposition: Given a fixed-point-less endomorphism on B there is no surjective morphism A→B^A.
Most famous is certainly Cantor's Diagonalization, which introduced the technique and founded the field of set theory. For this we work in the category of sets where morphisms are functions. Let A=ℕ and B=2={0,1}. Now the function 2→2, 0↦1, 1↦0 witnesses that there can not be a surjection ℕ→2^ℕ, and thus there is more than one infinite cardinal. Similarly it is also the prototypiacal proof of incompletness arguments, such as Gödels Incompleteness Theorem when applied to a Gödel-numbering, the Halting Problem when we enumerate all programs (more generally Rice's Theorem), Russells Paradox, the Liar Paradox and Tarski's Non-Defineability of Truth when we enumerate definable formulas or Curry's Paradox which shows lambda calculus is incompatible with the implication symbol (minimal logic) as well as many many more. As in the proof for Curry's Paradox it can be used to construct a fixed-point combinator. It also is the basis for forcing but this will be discussed in detail at a later date.
If we were to replace point-surjective with epimorphism the theorem would no longer hold for general categories. (Of course in Set the epimorphisms are exactly the surjective functions.) The standard counterexample is somewhat technical and uses an epimorphism ℕ→S^ℕ in the category of compactly generated Hausdorff spaces. This either made it very obvious to you or not at all. Either way, don't linger on this for too long. (Maybe in future installments we will talk about Polish spaces, then you may want to look at this again.) If you really want to you can read more in the nLab page mentioned below.
This proof requires our category to be cartesian closed. This means that it has all finite products and gives us some "meta knowledge", called closed monoidal structure, to work with exponentials.
Yanofsky's theorem is a slight generalization. It combines our proof steps where we use the closed monoidal structure such that we only use finite products by pre-evaluating everything. But this in turn requires us to introduce a corresponding technicallity to the statement of the theorem which makes working with it much more cumbersome. So it is worth keeping in the back of your mind that it exists, but usually you want to be working with Lawvere's version.
Yes you can. No, you will never be able to look at this diagram the same way again.
We see that Lawvere's Theorem forms the foundation of foundational mathematics and logic, appears everywhere and is (imo) its most important theorem. Hence why I thought it a good pick to kick of this series.
If you want to read more, the nLab page expands on some of the only tangentially mentioned topics, but in my opinion this suprisingly beginner friendly paper by Yanofsky is the best way to read about the topic.
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