A cartesian closed category is pointed if and only if it has a zero object

Question

I am solving the following problem: Let $\mathbf{A}$ be a cartesian closed category (i.e. it has finite products and has exponentials). Then TFAE:

(i) $\mathbf{A}$ pointed(i.e. for all $A,B \in Ob\mathbf{A}$ there is a zero morphism $z: A \rightarrow B$ where a morphism is a zero morphism iff it's both constant and co-constant where $z:A\rightarrow B$ is constant iff for all $f,g: C\rightarrow A$ : $z\circ f = z \circ g$.

(ii) $\mathbf{A}$ has a zero object $0$.

(iii) $\mathbf{A}$ is equivalent to $\mathbf{1}$.

I have proven that $(iii)\Rightarrow (ii) \Rightarrow (i)$ and $(ii) \Rightarrow (iii)$. I am struggling with the implication $(i) \Rightarrow (ii)$ and/or $(i)\Rightarrow (iii).$ I thought of showing that for each object $A$, $A$ is an initial object or if $T$ is a terminal object, then $T$ is a zero object.

Because if $0$ is a zero-object, then $0 \simeq A\times 0 \simeq A$. On the other hand if $A$ is an initial object for each $\mathbf{A}$ object, then so is $T$ thus $T \simeq I$ and we are done.

Is there anybody that can provide me with a hint, because I am sure I am missing something ridiculous...

Answer 1

Assume that $\mathbf{A}$ is pointed with a terminal object $*$. We want to show that $\hom(*,X)$ has only one element for all $X \in \mathbf{A}$, i.e. $*$ is also initial (this will show $(i) \implies (ii)$). By pointedness, there's a morphism $0_X : * \to X$ such that for all $f,g : X \to Y$, $f \circ 0_X = g \circ 0_X$. For $X = *$, this morphism $0_*$ has to be the identity $\operatorname{id}_*$, for there is only one morphism $* \to *$ ($*$ is terminal).

Hence if $f : * \to X$ is some other morphism (for an arbitrary $X$), then $$f = f \circ \operatorname{id}_* = f \circ 0_* = 0_X \circ 0_* = 0_X \circ \operatorname{id}_* = 0_X.$$

This proves that $*$ is in fact initial: $\hom(*,X) = \{0_X\}$.