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In a category with zero morphisms, can someone think of an example where $A\rightarrow B$ is a zero monomorphism but $A$ is not a zero object?

(It is easy to see that $A$ should be a terminal object.)

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By "category with zero morphisms" I will assume you mean a category enriched over pointed sets, so that there is a distinguished zero morphism between any two objects. Suppose $0_{A,B} : A \to B$ is a monomorphism and also a zero morphism. Then, $$0_{A,B} \circ \textrm{id}_A = 0_{A,B} \circ 0_{A,A}$$ and so cancelling $0_{A,B}$, we get $\textrm{id}_A = 0_{A,A}$. It follows that $A$ is a zero object.

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Consider the category of sets and let $A = \lbrace a \rbrace$. Certainly, the category of sets has zero morphisms. However, $A \to \lbrace a,b\rbrace$ is a zero morphism sending $a$ to $a$. But $A$ is not a zero object.

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    You are correct clive. I will edit my answer to reflect your correction. Thank you!2012-09-14
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    Sorry, I keep mis-reading the post. This now works; ignore my previous comments!2012-09-14
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    It would be interesting to have an example of a category with zero objects answering ashpool's question. However, as stated, the category only needs zero morphism.2012-09-14
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    Not really. A zero object is both a terminal object _and_ an initial object, by definition.2012-09-14
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    What I meant by "category with zero morphisms" is that the category has a zero morphism between every pair of objects satisfying certain conditions (composition with a zero morphism is a zero morphism). As such, I don't think the category of sets qualifies.2012-09-14
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    @ashpool: correct me if I'm mistaken, but in the category of sets we have zero morphism between *any* two objects other than empty set (we don't have any morphisms to empty set except identity...) -- any constant function works.2012-09-14
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    @tomasz: A constant function between sets is not necessarily coconstant. Take $f:A\rightarrow B$ where $A\neq\emptyset$ and $|B|>1$.2012-12-02
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    @ashpool: It isn't? Could you provide a more concrete example?2012-12-02
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    [Wikipedia][1] says that a coconstant map f is such that $\forall\ g,h$ that can be composed on the right of f, $g\circ f = h\circ f$. With $f$ constant, that means that g and h evaluated on the image of f are equal. That is not true if $A\neq \emptyset$ [1]: https://en.wikipedia.org/wiki/Zero_morphism2014-06-14
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If your category contains zero objects, then all terminal objects are initial, and hence zero objects. (And dually, all initial objects are terminal.)

Proof: Let $T$ be a terminal object, $Z$ be a zero object and $X$ be any object. There are unique arrows $T \overset{i}{\underset{j}{\rightleftarrows}} Z$ and these must form an isomorphism since the identities are the unique arrows $T \to T$ and $Z \to Z$. If $X$ is any object then there is a unique arrow $f:Z \to X$, but then there is an arrow $fi:T \to X$, and if $g:T \to X$ is another such arrow then $f=gj$ and so $fi=gji=g$. Hence $T$ is initial. $\square$

So, to half-answer your question, if $A \to B$ is a zero monomorphism in a category $\mathcal{C}$ then, since $A$ is necessarily terminal, $A$ is a zero object if and only if $\mathcal{C}$ has a zero object. So to find the example you seek, all you have to do is find a zero monomorphism in a category without a zero object.