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 $\operatorname {\mathrm{Obj}}(\mathcal{C})=\left\{ A,B\right\} $.

• •

  Morphisms. We have

  \begin{align*} \operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,A) & = \left\{ e_{A},\operatorname {\mathrm{id}}_{A}\right\} ,\\ \operatorname {\mathrm{Hom}}_{\mathcal{C}}(B,B) & = \left\{ e_{B},\operatorname {\mathrm{id}}_{B}\right\} ,\\ \operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,B) & = \left\{ f,g\right\} ,\\ \operatorname {\mathrm{Hom}}_{\mathcal{C}}(B,A) & = \text{Ø}. \end{align*}
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  Composition. The nontrivial compositions in $\mathcal{C}$ are the following:

  \[ \begin{aligned} e_{A}\circ e_{A} & = \operatorname {\mathrm{id}}_{A},\\ e_{B}\circ e_{B} & = \operatorname {\mathrm{id}}_{B}, \end{aligned} \quad \begin{aligned} f\circ e_{A} & = g,\\ g\circ e_{A} & = f, \end{aligned} \quad \begin{aligned} e_{B}\circ f & = f,\\ e_{B}\circ g & = g. \end{aligned} \]

We may picture $\mathcal{C}$ as follows:

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 $\operatorname {\mathrm{Nat}}(\iota _{A},\iota _{B})=\text{Ø}$: A natural transformation $\alpha \colon \iota _{A}\Rightarrow \iota _{B}$ consists of a morphism

  \[ \alpha \colon \underbrace{\iota _{A}(\bullet )}_{=A}\to \underbrace{\iota _{B}(\bullet )}_{=B} \]

  in $\mathcal{C}$ making the diagram

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

     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. (b)

     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 $\operatorname {\mathrm{Nat}}(F\circ \iota _{A},F\circ \iota _{B})\cong \mathrm{pt}$: A natural transformation

  \[ \beta \colon F\circ \iota _{A}\Rightarrow F\circ \iota _{B} \]

  consists of a morphism

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

  in $\mathbb {1}$ making the diagram

  
  commute for each $e\in \operatorname {\mathrm{Hom}}_{\mathsf{B}\mathbb {Z}_{/2}}(\bullet ,\bullet )\cong \mathbb {Z}_{/2}$. 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.

This finishes the proof.

is full. However, by Item 5 of Proposition 11.10.1.1.2, we have isomorphisms of categories

and the diagram

• •

  Item 7a and Item 7b Are Equivalent: This is true by the definition of corepresentably full morphism; see Chapter 14: Types of Morphisms in Bicategories, Definition 14.2.2.1.1.

• •

  Item 7a, Item 7c, and Item 7d Are Equivalent: See , of .

• •

  Item 7a and Item 7e Are Equivalent: See Item (b) of Remark 4.3 of [AESV, On Functors Which Are Lax Epimorphisms].

This finishes the proof.