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Dec
9
revised Proving invariance of $ds^2$ from the invariance of the speed of light
edited tags
Dec
9
revised Adjoint of forgetful functor
edited tags
Dec
9
comment When the homsets of a category are structured.
It is not enough to say that the hom-sets are structured: that in itself can happen for trivial reasons, e.g. sets equipped with a $G$-action where $G$ is a fixed group. So we must also relate the structure to the category somehow.
Dec
9
comment When the homsets of a category are structured.
If you have a commutative algebraic theory, then the category of models has a canonical monoidal closed structure.
Dec
9
comment Proof of the five lemma
@HagenvonEitzen The usual five lemma follows from the short five lemma: factor each morphism appearing in the rows into an epimorphism followed by a monomorphism.
Dec
8
comment Is an equivalence an adjunction?
Adjoints are unique up to unique isomorphism.
Dec
8
comment Hom-functor preserves pullbacks
It preserves all limits. Try showing that instead.
Dec
8
answered Does the functor $\mathbf{cosk_n}:sSet\to sSet$ preserve Kan complexes?
Dec
7
awarded  Nice Question
Dec
7
answered How do we get the canonical cokernel-kernel decomposition in a pre-abelian category?
Dec
6
answered Is an equivalence an adjunction?
Dec
6
comment Why is the colimit over this filtered index category the object $F(i_0)$?
Yes. But then that means $\mathcal{I}$ has a terminal object.
Dec
5
revised Why is the colimit over this filtered index category the object $F(i_0)$?
added 209 characters in body
Dec
5
revised Why is the colimit over this filtered index category the object $F(i_0)$?
added 460 characters in body
Dec
5
answered Why is the colimit over this filtered index category the object $F(i_0)$?
Dec
5
revised How Co-Comma Categories are constructed?
deleted 8 characters in body
Dec
5
comment In type theory, why isn't $x = x' : X$ simply wrong?
This is part of the syntax of type theory. Note that this is not a proposition, so it does not have a truth value.
Dec
4
answered Is the map into the terminal object an epimorphism?
Dec
4
comment Can the ultrafilters in the poset of open subsets be made into a topological space?
A topological space is sober if and only if it is homeomorphic to its soberification.
Dec
4
comment Can the ultrafilters in the poset of open subsets be made into a topological space?
This is called the soberification of $X$.