This week’s Google Hangout (RSVP here) will cover problems/questions from week ten of the (now revised) syllabus:

- 5.1 Categories and functors
- 5.2 Common categories and functors from pure math

Last week we made it through most of section 5.1, so we’ll finish it up and hopefully make it through 5.2, and pending time and people’s motivation this past week, possibly break into the beginning of 5.3.

We’re behind our original schedule, but still making some relatively good progress with a core peleton of people. Given the participation, we’ve slowed down a tad in an effort to not lose anyone, particularly with a variety of schedules and time zones.

Now that we’re into chapter 5 where the real fun has begun, people might also consider branching out to some of the alternative and more advanced texts. Some of the video resources are probably more germane to everyone now as well. In particular, I spent some time last week with Dr. Codrington’s youtube videos (1-6 of his “Lesson One”) which are generally excellent and which follow reasonably enough the presentation of Spivak that one won’t become lost.

Again, as a reminder for those who have had difficulties joining in the weekly live conversation, be sure to log into Google+ a few minutes before the start of the meeting. Once it starts the moderators send out additional invitation reminders to join the conversation which should result in a pop up on your Google+ page inviting you into the live video/audio feed.

Alternately you can go to the Category Theory Study Group’s Google+ posts page where you should see a button with a camera icon on the particular week’s post that says “Join Hangout”. Clicking it should put you into the live conversation.

Another option should be to join from your events page. If you don’t see the hangout event there, make sure we have your G+ account so we can ensure you’re included in the invitation.

]]>f(0) = 3

In FLin[2] 0 <= 1. So f(0) <= f(1). If f(0) is 3 then f(1) and f(2) has to be 3 for f(0) <= f(1) and f(0) <= f(2) and f(1) <= f(2). Basically we need to find 3 non decreasing elements in FLin[3]. Rest is combinatorics.

]]>B. for every pair *x*, *y* ∈ Ob(C), a set HomC(x,y)∈Set; it is called the *hom-set from x to y*; its elements are called *morphisms from x to y*;^{2}

Footnote 2 says

The reason for the notation Hom and the word *hom-set* is that morphisms are often called *homomorphisms*, e.g., in group theory.

But morphisms and homomorphisms are different, right ? Morphism can be anything, homomorphisms are more restrictive. Is that correct ? For eg: there can be morphisms between 2 groups that are not homomorphisms, right ?

]]>Ex. 5.2.1.26 asks

How many functors are there of the form **GrIn** → **SGrIn**?

I did not understand the answer and explanation given in the textbook. Can anyone please explain ?

]]>A. there is a function U:Ob(Q)→Ob(C),

B. for all a,b∈Ob(Q), we have an injection U:HomQ(a,b)↪HomC(U(a),U(b)),

Why is the function and injection both called U ? Shouldn’t the injection be called something other than U ? This is more a question of naming but I just wanted to confirm.

]]>This week’s Google Hangout (RSVP here) will cover problems/questions from week seven of the syllabus:

- 4.4 Orders
- 4.5 Databases: schemas and instances

If you’re joining us in progress, please feel free to add in any questions you might have about previous material as well – it’s never too late to join us all.

If you’re stuck and can’t make it, it will be archived (barring any further technical difficulties) on our YouTube Channel for later consumption.

]]>A graph (*V*, *A*, *src*, *tgt*) involves two sets and two functions. For two graphs to be comparable, their two sets and their two functions should be appropriately comparable. Let *G* = (*V*, *A*, *src*, *tgt*) and *G*′ = (*V*′, *A*′, *src*′, *tgt*′) be graphs. A *graph homomorphism f from G to G*′, denoted *f* : *G* → *G*′, consists of two functions *f*_{0}: *V* → *V*′ and *f*_{1}: *A* → *A*′ …the rest…

So this means that not all elements of A’ need to have a mapping from A and not all elements of V’ need to have a mapping from V. For example if G has m vertices and n edges and G’ has m'(>1) vertices and n'(>1) edges we could have all vertices of G go to one vertex of G'(say a) and all edges of G go to an edge from a to a(assuming that exists) in G’. That would be a valid homomorphism. Am I understanding that correctly ?

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This week’s Google Hangout (RSVP here) will cover problems/questions from week six of the syllabus:

- 4.3 Graphs

If you’re joining us in progress, please feel free to add in any questions you might have about previous material as well – it’s never too late to join us all.

If you’re stuck and can’t make it, it will be archived (barring any further technical difficulties) on our YouTube Channel for later consumption.

For those who have had difficulties joining in the weekly live conversation, be sure to log into Google+ a few minutes before the start of the meeting. Once it starts the moderators send out additional invitation reminders to join the conversation which should result in a pop up on your Google+ page inviting you into the live video/audio feed.

Alternately you can go to the Category Theory Study Group’s Google+ posts page where you should see a button with a camera icon on the particular week’s post that says “Join Hangout”. Clicking it should put you into the live conversation.

Another option should be to join from your events page.

Some will notice that we’ve slowed down a tad to accommodate the peleton who have been doing the diligence to keep up. Hopefully the last week “off” for the July 4th holiday in the states will have let everyone catch up a bit. The syllabus will be changing to meet our needs as we proceed.

For those with some general background in Category Theory, we’ll be getting into the serious material shortly.

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