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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...

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  • $\begingroup$ Any terminal object in a pointed category is a zero object. $\endgroup$ – Zhen Lin Feb 18 '16 at 15:06
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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: $\hom(*,X) = \{0_X\}$.

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  • $\begingroup$ Thanks, now I feel stupid... :) $\endgroup$ – sqtrat Feb 18 '16 at 15:17
  • $\begingroup$ @sqtrat Don't, it just takes practice to solve these exercises, not being able to solve them doesn't make you stupid. You were able to prove three out of the four implications on your own! $\endgroup$ – Najib Idrissi Feb 18 '16 at 15:26
  • $\begingroup$ Thanks for your time anyway, I just got a little frustrated with not being able to solve it, that's all. $\endgroup$ – sqtrat Feb 18 '16 at 15:37

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