A functor \(\mathcal{C}\xrightarrow{F}\mathcal{D}\) between two categories.
For each object in \(\mathcal{C}\) one specifies \(F(c) \in Ob(\mathcal{D})\)
For each morphism \(c_1\xrightarrow{f}c_2\) in \(\mathcal{C}\), one specifies \(F(c_1)\xrightarrow{F(f)}F(c_2)\) in \(\mathcal{D}\)
Furthermore, two properties must be satisfied:
Identity morphisms are mapped to identity morphisms
Composition is preserved: \(F(f;g)=F(f);F(g)\)
Between the free square category and the commutative square category, there is no functor from the commutative square category to the free square category which keeps the corners in the same place.
If we did this, we’d have \(F(f;h)=F(g;i)\) (since these are the same morphism).
The functor rules would allow us to break this up into \(F(f);F(h)=F(g);F(i)\) and we don’t have a choice for those mappings other than \(f;h=g;i\), something that is not true in the free square category.
Functors between preorders are monotone maps. Morphisms in the source must map to sources in the target, so if \(a \leq_P b\), then we require \(F(a) \leq_Q F(b)\), which is tantamount to the monotone map constraint.
How many functors are there from 2 to 3?
We are only concerned about where the objects go, since the target category is a thin category (there is no choice about which morphism is mapped to). Thus the functors are: 11, 22, 33, 12, 23, 13
Say where each of the 10 morphisms in the free square category are mapped to by a functor to the commutative square category (where \(Ob(F)\) maps each corner to the same corner).
The four identity morphisms and four length-1 paths are trivially mapped to the the corresponding morphisms. Both length-2 paths in the free category are mapped to the same morphism, \(f;h\).
Consider \(\mathcal{C}=\boxed{\bullet \rightarrow \bullet}\) and \(\mathcal{D}=\boxed{\bullet \rightrightarrows \bullet}\). Give two functors that act the same on objects but differently on morphisms.
Given a category \(\mathcal{C}\), show that there exists a functor \(id_\mathcal{C}\) known as the identity functor on \(\mathcal{C}\)
Show that given \(\mathcal{C}\xrightarrow{F}\mathcal{D}\) and \(\mathcal{D}\xrightarrow{G}\mathcal{E}\) we can define a new functor \(\mathcal{C}\xrightarrow{F;G}\mathcal{E}\) just by composing functions.
Show that there is a category, let’s call it Cat where the objects are categories, morphisms are functors, and identities/composition are given as above.
Mapping objects and morphisms to themselves satsifies the functor constraints of preserving identities and composition.
If \(F,G\) both independently preserve identity arrows, then composition of these will also preserve this. Also \(G(F(f;g))=G(F(f);F(g))=G(F(f));G(F(g))\) using the independent facts that \(F,G\) each preserve composition.
Composition of identity functions do not change anything, so the identity functor (defined by identity function) will obey unitality. Because function composition is associative and functor composition is defined by this, we also satisfy that constraint.