Tag Info

30

Well. That depends on whom you might ask this. Set theory might be inconsistent. In particular $\sf ZFC$ and its extension by large cardinal axioms. It's a nontrivial thing, to feel safe with these theories, and it takes a lot of practice and time until you understand that $\sf ZFC$ is self-evident to some extent, and [some] large cardinal axioms are ...

16

No. The reason is that the splitting field is not unique up to unique isomorphism, and any terminal object has to be unique in this stronger sense.

7

If you're looking for motivation to pursue other foundations, I recommend the article "Rethinking Set Theory" by Tom Leinster: http://arxiv.org/pdf/1212.6543v1.pdf. In particular, he's providing a gentle, well-motivated introduction to Lawvere's Elementary Theory of the Category of Sets (ETCS). Leinster mentions at least two complaints. First, in ...

5

The categories are not even equivalent. In $\mathbf{Set}$ there exists an arrow from a terminal object to a non-terminal object -- for example one of the functions $\{1\}\to\{1,2\}$. But $\mathbf{Set}^{op}$ does not have this property, because such an arrow in $\mathbf{Set}^{op}$ would be a function from a nonempty set to the empty set.

4

It simply doesn't follow that $g$ is a morphism in $S$. For example, $A$ might be the category of abelian groups, $S$ might be the full subcategory of finitely generated abelian groups, and the target of $g$ might be an infinitely generated abelian group. But the proof is very easy to repair: just define $g = \text{coker}(f) \in A$ in the first place. Then ...

4

I contend that we can reconstruct $\mathbb Z_{>0}$ up to a permutation of the form $$\prod_i p_i^{a_i}\mapsto \prod_i \pi(p_i)^{a_i}.$$ Since such maps are homomorphisms of the multiplicative monoid we can reconstruct multiplication. We can associate with every $x$ the set of proper divisors, i.e., $D(x):=\{\,y\in\mathbb Z_{>0}:y\ne x, y\mid x\,\}$. ...

3

Yes. In SET singletons are terminal objects. Then in its opposite category any singleton is an initial object. However in SET there is only one initial object (the empty set). This shows that the two categories are not isomorphic.

2

The question reduces to a question about sheaves of sets, so I will just work with those (instead of modules or whatever). The point is that we have the following commutative diagram of right adjoint functors, $$\require{AMScd} \begin{CD} \mathbf{Sh}(X) @>>> \mathbf{Sh}(Y) \\ @VVV @VVV \\ \mathbf{Psh}(X) @>>> \mathbf{Psh}(Y) \end{CD}$$ ...

2

A mapping $f:X\to X$, for a set $X$, which satisfy $f\circ f=1$ is called involution.

2

Theorem: $(X_1 \times X_2) \times X_3$ together with the obvious projection arrows forms a ternary product of $X_1, X_2, X_3$. Proof Assume $[X_1 \times X_2, \pi_1, \pi_2]$ is a product of $X_1$ with $X_2$, and $[(X_1 \times X_2) \times X_3, \rho_1, \rho_2]$ is a product of $X_1 \times X_2$ with $X_3$. Take any object $S$ and arrows $f_i\colon S \to ... 2 If somebody referred to a full subcategory of an abelian category closed under extensions, then I'd assume that they probably meant the subcategory to be additive, and therefore containing a zero object. But you could, for example, take$\mathcal{A}$to be the category of finite dimensional vector spaces over a field, and$\mathcal{B}$to be the full ... 2 Normally this kind of thing is verified by the "what else?" argument, that is, what are the chances you really have two different canonical maps in that square? That said, it could be comforting to know it's possible to really check such a thing. So: Let me change your$\oplus_i$to$\oplus_j$, to avoid collision of notation. We have to show that for every ... 2 You don't need to use the fact that$φ$is monomorphism any more - you already did that by assuming$Y → Y ×_Z Y$is isomorphism, which is equivalent to$φ$being mono. All that is left to do is to observe that the bottom square is also cartesian (for trivial reasons), which makes the whole outer square cartesian too. Now$g$, being a pullback of an ... 1 Here is a possible starting point.$\text{Grp}$, as well as other familiar categories of algebraic objects like$\text{Vect}$or$\text{Ring}$, are distinguished from arbitrary categories by the fact that they are categories of models of Lawvere theories in$\text{Set}$. This is a categorical way of talking about universal algebra. A categorical ... 1 Let$C$be any category whatsoever, and let$c \in C$be an object. I claim that the representable functor$\text{Hom}(c, -) : C \to \text{Set}$has a left adjoint iff all coproducts$\bigsqcup_S c$exist, and that when this is true the left adjoint sends a set$S$to$\bigsqcup_S c$. Theorem 3.4.5 guarantees that the$C$you care about is cocomplete, so in ... 1 First example. Let$\mathrm{Mor}(x,y)$be empty for$x<y$, and consists of one element if$x\ge y$. So, if you can write such a diagram,$(a_n)$is non-decreasing sequence. Therefore this diagram has the inverse limit iff$(a_n)$bounded above, and inverse limit equals to supremum. Second example. Let$\mathrm{Mor}(x,y)$be the set of all paths ... 1 We have answer using projections to prove universal property by Peter Smith, and I'm going to show another way to prove this, although, depending how one defines products (limits), this may actually be longer to work out from scratch. But, the characterization of product I'm going to use is very useful in itself:$P\$ is product of the family ...

Only top voted, non community-wiki answers of a minimum length are eligible