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Feb
1
answered Type theoretic proof that $\lnot (A \lor B) \Rightarrow \lnot A \land \lnot B$
Feb
1
comment Example of Spherical Element (Simplicial Homotopy)
AAAAAAAH Sorry I forgot you have degeracy maps in $\Delta$, so stupid by me. Again apologize for the confusion.
Feb
1
comment Example of Spherical Element (Simplicial Homotopy)
@NajibIdrissi is the simplicial set with only one $0$-simplex and no simplex in higher dimension. Clearly this cannot be a the terminal simplicial set.
Feb
1
comment Example of Spherical Element (Simplicial Homotopy)
@NajibIdrissi if you like you could think them as two different kind of slice-categories one is the canonical $1/\mathbf{sSet}$, where $1$ is the terminal presheaf, the other one is $S^{0}/\mathbf{sSet}$, where $S^{0}$ represent the simplicial set where $S^{0}_0$ is a singleton and each $S^0_n$ is empty for $n > 0$.
Feb
1
comment Example of Spherical Element (Simplicial Homotopy)
I see, let me be more clear about my doubt. The problem lies in the notion of point you want to use: by point do you mean an element of $X_0$ (that is a $0$-simplex) or a morphism from $1$ (the terminal simplicial set) into $X$?
Feb
1
comment Example of Spherical Element (Simplicial Homotopy)
Could you point out your definition of pointed simplicial set? There are at least two that come to my mind.
Feb
1
comment Prove that a group G is finitely generated if and only if there is a surjective homomorphism $F(\{1,..,n \}) \to G$
@Jxt921 Ok. I wanted to give an hint.... but unfortunately I couldn't find any so I provided a solution. The point is that it seems quite straightforward once you know the properties of homomorphisms and the concrete characterization of free groups......
Feb
1
answered Prove that a group G is finitely generated if and only if there is a surjective homomorphism $F(\{1,..,n \}) \to G$
Feb
1
comment Prove that a group G is finitely generated if and only if there is a surjective homomorphism $F(\{1,..,n \}) \to G$
I don't understand, are you looking for a solution to your problem or do you want an hint?
Feb
1
revised Prove that a group G is finitely generated if and only if there is a surjective homomorphism $F(\{1,..,n \}) \to G$
Grammar fix
Jan
31
answered How to simulate power sets in structural set theory (ETCS)?
Jan
31
revised Defining natural transformations based on generalized elements?
Grammar corrections
Jan
31
comment Defining natural transformations based on generalized elements?
As an additional remark, yoneda also implies that the natural isomorphism $\langle -,-\rangle$ determines completely the structure of the product on $A \times B$.
Jan
31
answered Defining natural transformations based on generalized elements?
Jan
28
awarded  Nice Answer
Jan
28
comment If $F : \mathbf{C} \to \mathbf{D}$ is an equivalence, does $\alpha_{GD} = G(\beta_D)$ hold in general?
Finally if you look carefully to the counterexample above that exactly what it does: it takes two unrelated natural transformation between the identity functors, i.e. two elements of the set $\mathbb C^{\mathbb C}[1_{\mathbb C},1_{\mathbb C}]$. To prove that the functor $1_{\mathbb C}$ is self-equivalent we just need to provide an element of that set, this could be the identity but it doesn't have to be it.
Jan
28
comment If $F : \mathbf{C} \to \mathbf{D}$ is an equivalence, does $\alpha_{GD} = G(\beta_D)$ hold in general?
I think that the point is that while we can choose many different, unrelated, proofs of the fact that $F$ and $G$ are equivalences, if we want that our proofs also .... prove that the functors are adjoint we need to choose them more carefully.
Jan
28
comment If $F : \mathbf{C} \to \mathbf{D}$ is an equivalence, does $\alpha_{GD} = G(\beta_D)$ hold in general?
@StudentType think of the natural isos as proofs of the equivalences between the functors. An equivalence of categories provides two functors between the categories with two proofs that these functor are equivalences of categories. In an adjoint equivalence we need to require something more, for instance we want that the two proofs satisfy the triangle identities.
Jan
27
revised Do mathematical objects have underlying types?
Left unrelated tags
Jan
27
answered Quantifier notation: $\forall n \implies \cdot$ versus $\forall n, \cdot$