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4h
comment Inverse image sheaf functor: proving $(f^\ast P)_x=P_{f(x)}$
@Arrow The codomain category would be the (very large) category of (large) categories. It is literally a functor the way I define it.
12h
answered Inverse image sheaf functor: proving $(f^\ast P)_x=P_{f(x)}$
12h
comment Inverse image sheaf functor: proving $(f^\ast P)_x=P_{f(x)}$
The only abstract nonsense proof I am aware of is restricted to sheaves (not presheaves), and it is based on the fact that the composite of two left adjoints is the left adjoint of the composite (of the right adjoints).
1d
awarded  Popular Question
Jul
3
comment Intuitionistic Proof of $(a \Rightarrow b) \Rightarrow (\lnot b \Rightarrow \lnot a)$
@z5h No, the point is that $\lnot a$ is an abbreviation of $a \to \bot$, so if you want to prove $\lnot a$, you should assume $a$ and prove $\bot$; similarly, $\lnot b$ is an abbreviation of $b \to \bot$, so if you want to deduce $\bot$ from $\lnot b$, you should prove $b$.
Jul
3
answered Intuitionistic Proof of $(a \Rightarrow b) \Rightarrow (\lnot b \Rightarrow \lnot a)$
Jul
3
comment Distributivity of pullbacks
This is essentially the definition of a lextensive category.
Jul
3
awarded  Necromancer
Jul
3
comment Axiomatizability of the algebra of (a fragment of) calculus
I believe the answer is yes, because $S$ contains $\mathbb{Z} [x]$, and it seems to me that $\mathbb{Z} [x]$ is the free commutative ring equipped with a derivation $D$ and an element $x$ such that $D x = 1$.
Jul
3
comment Why is a simply connected 3-manifold a homotopy 3-sphere?
You should also include the part where triangularisability and Poincaré duality are explained.
Jul
2
comment Spaces $X$ and $Y$ with $[Z, X]_{\bullet} \cong [Z, Y]_{\bullet}$ for all cogroup objects $Z$ in $\mathsf{hTop}_{\bullet}$
Pointed topological spaces.
Jul
2
comment Properties preserved under equivalence of categories
Everything in that list is preserved by either a left adjoint or a right adjoint, and equivalences are simultaneously left and right adjoints.
Jul
1
comment Is there a special name for functors from a category C to a subcategory of C?
They are endofunctors.
Jul
1
comment Group action on a category
As I said, it is a special case of a pseudofunctor.
Jul
1
answered Category of Sets and Bag-valued functions
Jul
1
comment Spaces $X$ and $Y$ with $[Z, X]_{\bullet} \cong [Z, Y]_{\bullet}$ for all cogroup objects $Z$ in $\mathsf{hTop}_{\bullet}$
They are different. You should clarify which one you mean.
Jun
29
comment Group action on a category
Yes, it is a pseudofunctor.
Jun
29
answered Group action on a category
Jun
29
comment How to show that a map is finite
You could use the valuative criterion for properness, but I think it would be much easier to just use the standard definition of finite morphism in terms of finite algebras.
Jun
29
comment How to show that a map is finite
Well, quasi-finiteness is clear, right? So you just need to check that the map is proper.