# Morphims with unique domain

A morphism $m$ of a category has the following property:

No morphism (except of the identity morphism) of the category has codomain equal to the domain of $m$. In other words, $m$ cannot be composed on the right.

In fact, I have a category with many such special morphisms.

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That is impossible, as any object of the category has the identity morphism to itself. –  Zev Chonoles Jun 12 '12 at 17:02
@ZevChonoles: You are right. I need to decide what to do with identity morphisms. –  porton Jun 12 '12 at 17:04
@ZevChonoles: I now think that I would do well not to require the non-existence of the morphisms with such codomains, just never compose $m$ on the right (not because this is impossible but I just don't want to compose it on the right and don't want to define what such composition would produce). –  porton Jun 12 '12 at 17:07
@ZevChonoles: But this way, I need to prove existence of a category having defined only left composition of $m$ and right composition to be left arbitrary. –  porton Jun 12 '12 at 17:12

Why should such a morphism have a special name? This property is a property of the domain, it does not depend on the specific morphism. Maybe you are looking for emptiness of an object, it is quite similar, but this notion requires the object to be an initial object (that would explain why there are that many of those morphisms ;)) and allows some additional isomorphisms.

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Maybe you should work out the situation in categorical terms, so we can see what is going on. To me it just means, that in your category $\mathbf{C}$ there are objects $B$, $C$ (with $B \neq C$ necessarily) with $hom(B,C) \neq \emptyset$ such that $hom(A,B) = \emptyset$ for all objects $A \neq B$. Is that what you mean?

In that case maybe you should look into graph theory. If you look on the category as a graph this means, that the vertex $B$ has no predecessors And maybe this has a name.

However from a category theoretic point of view this is nothing special and hence I don't think there is a special name for it.

Looking on your comments I think you would do well to study category theory more deeply. A category 'without precomposition' doesn't make sense at all. If you arrive at situations like that, it is usually an indicator that you should rethink what you are doing.

Moreover if you put constrains on SOME morphisms from $hom(B,C)$ you leave category theory and what is called 'natural' ...

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From the (directed) graph theoretic perspective, the name "source" might be appropriate. A source in a directed graph is a vertex that is not the head (=target) of any edge. –  Dan Ramras Jun 14 '12 at 18:17