Evan Washington
@eewash.bsky.social
yields falsehood when preceded by his own quotation
well, he practices what he preaches
January 31, 2026 at 2:33 AM
well, he practices what he preaches
ok she's kinda valid
January 15, 2026 at 9:31 PM
ok she's kinda valid
i can smell this tweet from here
January 15, 2026 at 12:35 AM
i can smell this tweet from here
you all on here need to start posting half as good as this
December 30, 2025 at 9:04 AM
you all on here need to start posting half as good as this
amazing foreshadowing. all the villains were introduced way back in season 2
![tweet from Olivia Nuzzi on April 24, 2019: "Nothing but respect for MY president, @bariweiss" followed by a link to a Vanity Fair article about her. Bari Weiss replies: "Olivia! [heart eyes emojis]"](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:judtbidg6b2ye3jzverecw3g/bafkreidwo7666ejnzflryhzdfrde7xsiqooj55tnn34h3ymcvro6gtjnge@jpeg)
December 22, 2025 at 12:15 AM
amazing foreshadowing. all the villains were introduced way back in season 2
turns out they're both annoying writers www.telos.news/p/part-1-how...
November 18, 2025 at 1:08 AM
turns out they're both annoying writers www.telos.news/p/part-1-how...
what is the bisexual hand gesture
November 14, 2025 at 5:18 PM
what is the bisexual hand gesture
another old tweet so good I have to repost it here. frankly my talent is being wasted
November 2, 2025 at 9:10 PM
another old tweet so good I have to repost it here. frankly my talent is being wasted
a little over a year ago I asked for a nice notion of morphism between combinatorial games, and it looks like someone finally answered me with this paper! this seems neat
November 2, 2025 at 8:54 PM
a little over a year ago I asked for a nice notion of morphism between combinatorial games, and it looks like someone finally answered me with this paper! this seems neat
i've had nightmares less haunting than this pop-up
October 29, 2025 at 8:38 PM
i've had nightmares less haunting than this pop-up
so much to unpack in this cursed tweet
first of all...both sides of the aisle???
first of all...both sides of the aisle???
October 28, 2025 at 2:55 AM
so much to unpack in this cursed tweet
first of all...both sides of the aisle???
first of all...both sides of the aisle???
listen, there's a reason the Hyuuga clan doesn't have any Hokages
October 22, 2025 at 5:43 AM
listen, there's a reason the Hyuuga clan doesn't have any Hokages
smh Ace you can't be the Pirate King with a tattoo like that
October 22, 2025 at 5:39 AM
smh Ace you can't be the Pirate King with a tattoo like that
more women guards!!!!!
October 22, 2025 at 5:13 AM
more women guards!!!!!
you're kidding. no way. the margaret thatcher fan girl who wants to rebuild the ethnostate isn't necessarily a feminist? whoa
October 22, 2025 at 5:07 AM
you're kidding. no way. the margaret thatcher fan girl who wants to rebuild the ethnostate isn't necessarily a feminist? whoa
graphic design is my passion
October 20, 2025 at 8:50 PM
graphic design is my passion
must admit I'm feeling pretty owned by Mr. FCC
September 18, 2025 at 10:01 PM
must admit I'm feeling pretty owned by Mr. FCC
they're posting like crazy over there. it's like going to the zoo
September 16, 2025 at 8:53 PM
they're posting like crazy over there. it's like going to the zoo
I used to tweet so good
September 13, 2025 at 8:54 PM
I used to tweet so good
the only one of these that I actually haven't seen in the wild is (3) but still. seems like there's no obvious path
![What's the "dual" of a set $A$ with some operations $A^n \to A$ for some $n$?
1. **Comodels**: Of course, the dual of $A \times A$ must be $A + A$, so an $X$-ary function $A^X \to A$ has dual $$A \to X \cdot A$$ where $X \cdot A$ is just the $X$-fold coproduct $A + \dots + A$.
2. **Co-operations**: Of course, the dual of $A^2 = [2, A]$ must be $[A, 2] = 2^A$, so an $X$-ary function $A^X \to A$ has dual $$1^A \to X^A$$ which is, equivalently, an element of $X^A = [A, X]$.
3. **Co-structure of a signature:** For each $n < \omega$, let $F_n$ be the set of function symbols of arity $n$ in the signature $\Sigma$. Then a $\Sigma$-structure is a function $F_n \times A^n \to A$ for each $n < \omega$. Dually, a co-structure of $\Sigma$ should be a function $$A \to F_n \times A^n$$ for each $n < \omega$.
4. **Polynomial coalgebra**: A function $A \times A \to A$ is an algebra of the polynomial endofunctor $p: A \mapsto A^2$, so the answer should be a coalgebra of $p$, that is, a set $A$ and a function $A \to A \times A$. In general, we consider coalgebras of polynomial endofunctors, i.e. a set $A$ and a function $$A \to \sum_n X_n \cdot A^n$$](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:judtbidg6b2ye3jzverecw3g/bafkreidqsbqjxm5sxhl4byhcohbv3znavdihwqwz44uolhle6jguzazcom@jpeg)
September 12, 2025 at 10:18 PM
the only one of these that I actually haven't seen in the wild is (3) but still. seems like there's no obvious path