1.6. Initial and terminal objects

In this section we use notation and definitions from Leinster \cite{Leinster}.

Let $\cat{C}$ be a category. We call an $\objs{\cat{C}}$, $I\in\ob{\cat{C}}$ initial if for any object $C\in\ob{\cat{C}}$, there is exactly one morphism $f\in\ho[\cat{C}]{I}{C}$. We call an $\objs{\cat{C}}$, $T\in\ob{\cat{C}}$ terminal if for any object $C\in\ob{\cat{C}}$, there is exactly one morphism $f\in\ho[\cat{C}]{C}{T}$. The following Lemma \ref{lem:initialiso} and proof is adapted from Leinster \cite{Leinster} (page 49 Lemma 2.1.8)

Let $\cat{C}$ be a category and $I,I^\prime\in\ob{\cat{C}}$ initial objects in $\cat{C}$. There exists a unique isomorphism $f\in\ho[\cat{C}]{I}{I^\prime}$. Let $T,T^\prime\in\ob{\cat{C}}$ be terminal objects in $\cat{C}$ then there exists a unique isomorphism $g\in\ho[\cat{C}]{T}{T^\prime}$. \begin{proof} Since $I$ is initial there exists a unique morphism $f\in\ho[\cat{C}]{I}{I^\prime}$.

It remains to show $f$ is an isomorphism. Since $I^\prime$ is initial then there exists a unique morphism $f^\prime\in\ho[\cat{C}]{I^\prime}{I}$. Therefore $f^\prime\circ f\in\ho[\cat{C}]{I}{I}$ but since $I$ is initial $f^\prime\circ f$ is the unique morphism in $\ho[\cat{C}]{I}{I}$. Since $\cat{C}$ is a category we have $id_{I}\in\ho[\cat{C}]{I}{I}$, hence $id_{I}$ is the unique morphism in $\ho[\cat{C}]{I}{I}$ therefore,

\[f\circ f^\prime = id_{I}\]

similarly,

\[f^\prime\circ f = id_{I^\prime}\]

since $f\circ f^\prime\in\ho[\cat{C}]{I^\prime}{I^\prime}$ is unique. Therefore $f$ is an isomorphism.

Since $T$ is terminal there exists a unique morphism $g\in\ho[\cat{C}]{T^\prime}{T}$, also since $T^\prime$ is terminal there exists a unique morphism $g^\prime\ho[\cat{C}]{T}{T^\prime}$. Then we have,

\[g\circ g^\prime = id_{T^\prime}\]

since $g\circ g^\prime\in\ho[\cat{C}]{T^\prime}{T^\prime}$ is unique. Similarly, $$g^\prime\circ g = id_{T}$$ since $g^\prime\circ g\ho[\cat{C}]{T}{T}$ is unique. \end{proof}

Given a category there may or may not exist an initial or terminal objects but the above Lemma \ref{lem:initialiso} states that if there are initial objects they are all isomorphic and if there are terminal objects they are all isomorphic.

\subsection{Examples}

The following example can be found on nCatLab\cite{nLab}.

Let $\mathbf{Set}$ be the category of sets. The empty set $\varnothing\in\ob{\mathbf{Set}}$ is initial. Any singleton set $\{x\}\in\ob{\mathbf{Set}}$ is terminal. For any set $Y\in\ob{\mathbf{Set}}$ we have the unique map

\[\begin{align*} f\colon Y &\to \{x\},\\ y &\mapsto x. \end{align*}\]

hence $\{x\}$ is terminal. There also exists a unique function $\emptyset\colon \varnothing\to Y$ which is the empty function, hence $\varnothing$ is initial.

Let $\mathbf{Grp}$ be the category of groups. The trivial group $\{id\}$ is initial and terminal. Given any group $(G,\circ)$ There exists a group homomorphism,

\[\begin{align*} f\colon \{id\}&\to (G,\circ),\\ id&\mapsto id_{G}, \end{align*}\]

and also a group homomorphism,

\[\begin{align*} f\colon (G,\circ)&\to \{id\},\\ x&\mapsto id. \end{align*}\]

Both of these functions are unique since group homomorphisms must preserve identities.

\subsection{Initial and terminal objects as adjoint functors}

We call the category $\mathbf{1}$ with one object $1\in\ob{\mathbf{1}}$ and one morphism $id_{1}\in\ho[\mathbf{1}]{1}{1}$ the identity category.

The following Lemma \ref{lem:initermadj} is adapted from Leinster \cite{Leinster} (Page 49 Example 2.1.9)

Let $\cat{C}$ be a category. Given $C\in\ob{\cat{C}}$, there exists a functor $\func{I_{C}}{\cat{C}}{\mathbf{1}}$ which maps $$1\mapsto C$$ and, $$id_{1}\mapsto id_{C}.$$ Further, there exists a functor $\func{T}{\mathbf{1}}{\cat{C}}$ where for $C\in\ob{\cat{C}}$, $$C\mapsto 1$$ and for $f\in\ho[\cat{C}]{A}{B}$, $$f\mapsto id_{1}.$$

\begin{proof} $I_{C}$ is a functor since,

\begin{align} I_{C}(id_{1}) &= id_{C} \ &= id_{I_{C}(1)} \end{align} and,

\[\begin{align*} I_{C}(id_{1}\circ id_{1}) &= I_{C}(id_{1}) \\ &= id_{C} \\ &= id_{C}\circ id_{C} \\ &= I_{C}(id_{1})\circ I_{C}(id_{1}). \end{align*}\]

$T$ is a functor since for all $C\in\ob{\cat{C}}$, $$T(id_{C})=id_{1} = id_{T(1)}$$ and for any $f\in\ho[\cat{C}]{A}{B}$ and $g\in\ho[\cat{C}]{B}{C}$,

\[\begin{align*} T(g\circ f) &= id_{1} \\ &= id_{1}\circ id_{1} \\ &= T(g)\circ T(f). \end{align*}\]

\end{proof}

Given a category $\cat{C}$ let $I$ be the set of all functors of the form $I_{C}$ as in Lemma \ref{lem:funcadj}, there exists a bijection

\[\begin{align*} \psi\colon\ob{\cat{C}}&\to I,\\ C&\mapsto I_{C}. \end{align*}\]

\begin{proof} Given $I_{C}\in I$ there exists $C\in\ob{\cat{C}}$ such that $C\mapsto I_{C}$ by Lemma \ref{lem:funcadj}, hence $\psi$ is surjective.

Given $I_{C}=I_{C^\prime}$ then $C=C^\prime$ by Lemma \ref{lem:funcadj} hence $\psi$ is injective.

Therefore $\psi$ is a bijection. \end{proof}

Let $\cat{C}$ be a category and $C\in\cat{C}$ then: \begin{enumerate} \item $\func{I_{C}}{\mathbf{1}}{\cat{C}}$ is left adjoint if and only if $C$ is an initial object of $\cat{C}$; \item $\func{I_{C}}{\mathbf{1}}{\cat{C}}$ is right adjoint if and only if $C$ is a terminal object of $\cat{C}$. \end{enumerate} \begin{proof} (1): Let $I_{C}$ be a left adjoint. Then there exists an adjoint situation defined in Theorem \ref{thm:adjsituation}. Hence there is a unique functor $\func{T^*}{\cat{C}}{\mathbf{1}}$ but since the only functor from $\cat{C}$ to $\mathbf{1}$ is the one defined in Lemma \ref{lem:funcadj} we have $T^* = T$. Therefore, for each $C^\prime\in\ob{\cat{C}}$,

\[\begin{align*} \ho[\cat{C}]{I_{C}(1)}{C^\prime}&\cong \ho[\mathbf{1}]{1}{T(C^\prime)}\\ \implies \ho[\cat{C}]{C}{C^\prime}&\cong \ho[\mathbf{1}]{1}{1}\\ \implies \ho[\cat{C}]{C}{C^\prime}&\cong \{id_{1}\}. \end{align*}\]

Therefore $C$ is an initial object.

Let $C\in\ob{\cat{C}}$ be an initial object, therefore we have for any $C^\prime \in\ob{\cat{C}}$ $$\ho[\cat{C}]{C}{C^\prime}.$$ has one element. Then we have,

\[\begin{align*} &\ho{C}{C^\prime}\cong \{id_{1}\}\\ \implies& \ho{C}{C^\prime}\cong \ho[\mathbf{1}]{1}{1}\\ \implies &\ho[\cat{C}]{I_{C}(1)}{C^\prime}\cong \ho[\mathbf{1}]{1}{T(C^\prime)}. \end{align*}\]

Therefore by \ref{thm:adjsituation} $I_{C}$ is a left adjoint.

(2): Let $I_{C}$ be a right adjoint. Then there exists an adjoint situation defined in Theorem \ref{thm:adjsituation}. Hence there is a unique functor $\func{T^*}{\cat{C}}{\mathbf{1}}$ but since the only functor from $\cat{C}$ to $\mathbf{1}$ is the one defined in Lemma \ref{lem:funcadj} we have $T^* = T$. Therefore, for each $C^\prime\in\ob{\cat{C}}$,

\[\begin{align*} &\ho[\mathbf{1}]{T(C^\prime)}{1}\cong \ho[\cat{C}]{C^\prime}{I_{C}(1)},\\ \implies &\ho[\mathbf{1}]{1}{1}\cong \ho[\cat{C}]{C^\prime}{C},\\ \implies & \{id_{1}\} \cong \ho[\cat{C}]{C^\prime}{C}. \end{align*}\]

Therefore $C$ is a terminal object. Let $C\in\ob{\cat{C}}$ be a terminal object, therefore we have for any $C^\prime \in\ob{\cat{C}}$ $$\ho[\cat{C}]{C^\prime}{C}.$$ has one element. Then we have,

\[\begin{align*} &\{id_{1}\} \cong \ho[\cat{C}]{C^\prime}{C}\\ \implies& \ho[\mathbf{1}]{1}{1}\cong \ho[\cat{C}]{C^\prime}{C}\\ \implies& \ho[\mathbf{1}]{T(C^\prime)}{1}\cong \ho[\cat{C}]{C^\prime}{I_{C}(1)}. \end{align*}\]

Therefore by \ref{thm:adjsituation} $I_{C}$ is a right adjoint. \end{proof}

Let $\mathbf{Set}$ be the category of sets. We have the functor,

\[\begin{align*} \func{I_{\varnothing}}{\mathbf{1}&}{\mathbf{Set}}, \\ 1&\mapsto \varnothing. \end{align*}\]

Then for $X\in\ob{\mathbf{Set}}$, we have a universal morphism $(\eta_{X},C_{X})$, where $C_{X}=1$ and $\eta_{X}=\emptyset$, the empty function:

\[\begin{align*} \eta_{X}\colon X \mapsto \varnothing \end{align*}\]

\begin{equation*}

.\ \end{equation*} Therefore, $I_{\varnothing}$ is right adjoint.

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