The Clowder Project

defined as the composition

is a composition of surjective functions, it follows from that it is also surjective. Therefore $G\circ F$ is full.

• Objects. We have $\text{Obj}\webleft (\mathcal{C}\webright )=\webleft\{ A,B\webright\} $.
• Morphisms. We have
  \begin{align*} \textup{Hom}_{\mathcal{C}}\webleft (A,A\webright ) & = \webleft\{ e_{A},\text{id}_{A}\webright\} ,\\ \textup{Hom}_{\mathcal{C}}\webleft (B,B\webright ) & = \webleft\{ e_{B},\text{id}_{B}\webright\} ,\\ \textup{Hom}_{\mathcal{C}}\webleft (A,B\webright ) & = \webleft\{ f,g\webright\} ,\\ \textup{Hom}_{\mathcal{C}}\webleft (B,A\webright ) & = \text{Ø}. \end{align*}

• Composition. The nontrivial compositions in $\mathcal{C}$ are the following:
  \[ \begin{aligned} e_{A}\circ e_{A} & = \text{id}_{A},\\ e_{B}\circ e_{B} & = \text{id}_{B}, \end{aligned} \quad \begin{aligned} f\circ e_{A} & = g,\\ g\circ e_{A} & = f, \end{aligned} \quad \begin{aligned} e_{B}\circ f & = f,\\ e_{B}\circ g & = g. \end{aligned} \]

Next, let $\mathcal{D}$ be the walking arrow category $\mathbb {1}$ of Definition 9.2.5.1.1 and let $F\colon \mathcal{C}\to \mathbb {1}$ be the functor given on objects by

and on non-identity morphisms by

Finally, let $\mathcal{X}=\mathsf{B}\mathbb {Z}_{/2}$ be the walking involution and let $\iota _{A},\iota _{B}\colon \mathsf{B}\mathbb {Z}_{/2}\rightrightarrows \mathcal{C}$ be the inclusion functors from $\mathsf{B}\mathbb {Z}_{/2}$ to $\mathcal{C}$ with

Since every morphism in $\mathbb {1}$ has a preimage in $\mathcal{C}$ by $F$, the functor $F$ is full. Now, for $F_{*}$ to be full, the map

so this is impossible:

• Proof of $\text{Nat}\webleft (\iota _{A},\iota _{B}\webright )=\text{Ø}$: A natural transformation $\alpha \colon \iota _{A}\Rightarrow \iota _{B}$ consists of a morphism
  \[ \alpha \colon \underbrace{\iota _{A}\webleft (\bullet \webright )}_{=A}\to \underbrace{\iota _{B}\webleft (\bullet \webright )}_{=B} \]

  in $\mathcal{C}$ making the diagram

  

  commute for each $e\in \textup{Hom}_{\mathsf{B}\mathbb {Z}_{/2}}\webleft (\bullet ,\bullet \webright )\cong \mathbb {Z}_{/2}$. We have two cases:

  1. If $\alpha =f$, the naturality diagram for the unique nonidentity element of $\mathbb {Z}_{/2}$ is given by
     

     However, $e_{B}\circ f=f$ and $f\circ e_{A}=g$, so this diagram does not commute.

  2. If $\alpha =g$, the naturality diagram for the unique nonidentity element of $\mathbb {Z}_{/2}$ is given by
     

     However, $e_{B}\circ g=g$ and $g\circ e_{A}=f$, so this diagram does not commute.

  As a result, there are no natural transformations from $\iota _{A}$ to $\iota _{B}$.

• Proof of $\text{Nat}\webleft (F\circ \iota _{A},F\circ \iota _{B}\webright )\cong \text{pt}$: A natural transformation
  \[ \beta \colon F\circ \iota _{A}\Rightarrow F\circ \iota _{B} \]

  consists of a morphism

  \[ \beta \colon \underbrace{\webleft [F\circ \iota _{A}\webright ]\webleft (\bullet \webright )}_{=0}\to \underbrace{\webleft [F\circ \iota _{B}\webright ]\webleft (\bullet \webright )}_{=1} \]

  in $\mathbb {1}$ making the diagram

  

  commute for each $e\in \textup{Hom}_{\mathsf{B}\mathbb {Z}_{/2}}\webleft (\bullet ,\bullet \webright )\cong \mathbb {Z}_{/2}$. Since the only morphism from $0$ to $1$ in $\mathbb {1}$ is $f_{01}$, we must have $\beta =f_{01}$ if such a transformation were to exist, and in fact it indeed does, as in this case the naturality diagram above becomes

  

  for each $e\in \mathbb {Z}_{/2}$, and this diagram indeed commutes, making $\beta $ into a natural transformation.

is full. However, by Item 5 of Proposition 9.10.1.1.2, we have isomorphisms of categories

and the diagram

commutes. It then follows from Item 1 that $F$ is full.

• Item (a) and Item (b) Are Equivalent: This is true by the definition of corepresentably full morphism; see Chapter 11: Graphs, Definition 11.2.2.1.1.
• Item (a), Item (c), and Item (d) Are Equivalent: See of .
• Item (a) and Item (e) Are Equivalent: See Item (b) of Remark 4.3 of [Adámek–Bashir–Sobral–Velebil, On Functors Which Are Lax Epimorphisms].