ct.category Theory – Closure of presentable objects within finite limits

In a locally presentable category $ cal E $, there are arbitrarily large regular cardinals $ lambda $ such as $ lambda $-presentable (a.k.a. $ lambda $-compact) are closed under withdrawals. Namely, the withdrawal functor $ { cal E} ^ {( to leftarrow)} to cal E $ is a right assistant, so accessible. So, he keeps $ lambda $-presentable objects for arbitrarily large $ lambda $so just check that the $ lambda $objects -presentable in $ { cal E} ^ {( to leftarrow)} $ are the ones that are punctual in $ cal E $. (A version of this argument is given in this answer in the case of finished products.)

Of course, "arbitrarily big" means that for any cardinal $ mu $ there is a regular cardinal $ lambda> mu $ with this property. A stronger statement would be that this is true for all big enough regular cardinals $ lambda $, that is, invert the quantifiers and say that there is a $ mu $ such as all regular cardinals $ lambda> mu $ have this property $ lambda $-The presentable objects are closed under withdrawals). Is this stronger affirmation true?

Note that this is certainly not true all regular cardinals $ lambda $ have this property; counterexamples can be found here.

abstract algebra – For $ Omega $ an algebraic closure of k $, why is $ k[x]/ (f) otimes_k Omega cong Omega ^ m $?

Fix a field k $ and an algebraic closure $ Omega $ of k $. For a monic, irreducible, separable polynomial $ f in k[x]$, let $ f $ to divide $ Omega $ as

$$ f (x) = prod_ {i = 1} ^ m (x – a_i). $$

It was said that the following holds:

$$ frac {k[x]} {(f)} otimes_k Omega con prod_ {i = 1} ^ m frac { Omega[x]} {(x-a_i)} cong Omega ^ m. $$

I can see the Chinese rest theorem here, but what is not clear to me is the resolution of the tensor because I am not yet comfortable working with them. Why is the first isomorphism true?

I like Google Plus, but its closure is imminent. Sigh

Google Plus is one of my favorites because it sells your privacy, but they are out of business soon. Sigh…

Google Chrome does not force closure and keeps the raise – MacBook

I was trying to force Google Chrome to quit, I realize that there is a malware problem, but I still do not know what to do!

Can someone help me, please!

Thank you

Review of the closure of the service | Promotion Forum

Hi guys,

As you all know, we have a hard time providing an adequate exam service for many years. We have faced many problems, and one of the great changes we have made over the past few years has been the implementation of the community feedback structure. In the future, that's what we want to focus on. How are we going to put in place a system based solely on community feedback, and then, how are we going to offer incentives to users who choose to take their time to comment?

That said, we will have to make a lot of changes. In order for us to begin these changes, we close our review service. Our review team will no longer provide official website reviews. I encourage you still to use the option of community comments on this forum and I hope that many comments will be exchanged. Thank you for your understanding and cooperation.

Regards,
Cameron

[ Politics ] Open question: do you like the federal closure?

It makes me happy. That makes me smile. Most of these workers are not needed and the positions must be eliminated from the workers who have resigned. I hope it will last much longer and that the EBT cards and section 8 are cut off. When riots and looting begin, we, the real Americans, can use the Second Amendment wisely to eliminate the parasites once and for all. .

Algebraic Geometry – How to understand the closure of a point?

I meet the generic point definition in Hartshorne as follows:

A generic point for an irreducible closed subset $ Z $ is a point $ P $ such as $ Z = ${$ overline {P} $}, or {$ overline {P} $} means the closure of the whole consisting of the set of points $ P $.

I do not quite understand the meaning of "the whole set of point $ P $Does this mean that the whole only includes one point? $ P $?
I've looked for some examples. In a ring spectrum, the closure of a point, which is a primordial ideal, is the set of all the prime ideals containing the point. So, in this case, the definition makes sense to me. Are there different examples that can explain the definition? Thank you!

What do you think of Nancy Pelosi's decision not to have a state of the Union during the closure of the government?

Some Democrats did not want to reelect Pelosi to the presidency. They thought that she was too old, that she represented the former Democratic Party. They wanted fresh blood.

But Pelosi is very very politically savvy! I think it was a brilliant shot! You want to close the government? Ok, we will stop your SOTU. Brilliant!

For years, Republicans have wanted to think that politics does not require any real skill. They elected GW Bush's president after serving half a governorship in a "weak governor state" where the governor is only a main character. And he was completely incompetent. They hired Trump based on nothing but attitude and self-confidence, and his administration is a slum. Pelosi would have made a better president than one or the other because she KNOWS WHAT IT DID!

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maui – How can I book to see the sunrise in Haleakala, if the closure of the federal government hampers online bookings?

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algorithms – calculating transitive closure

This question does not come from your homework, but rather from test preparation:

The calculation of the place-in is a calculation in which the algorithm does not need space beyond the output size (beyond
beyond a fixed space complexity). Show how, from a graph (E, V = G), the transitive closure can be calculated
in place in the calculation of G * = (V, E *)

I've tried to represent the graph with the help of an adjacency matrix. What is the next step?