Are we not doing meetings anymore ? I did not get invites for last week or this week.

# All posts by Arun Mathews

# Remark about 5.1.1.13

The author is asking about Homomorphism of FLin. In section 4.4.4 he talks about morphism of orders. If s1 <= s2 then f(s1) <= f(s2). Also, order is reflexive (s <= s). So when we have a morphism from FLin[2] to FLin[3] we need to pick 3 elements in FLin[3] so that they are related like they were in FLin[2]. The 3 elements we need to map are 0, 1, 2. If we map 0 in FLin[2] to 3 in FLin[3] then

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.

# Morphisms and homomorphisms

In section 5.1.1 where category is defined the author says

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 ?

# Graph indexing category – 5.2.1.26

In section 5.2.1.21 author talks about graph-indexing category(GrIn) and symmetric graph-indexing category(SGrIn).

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 ?

# Question about Questionable Category (Remark 5.1.1.6 Book)

In the remark author says:

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.

# Question about graph homomorphism

According to the definition:

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 ?