# “orthogonal” and “parallel” morphisms?

Let $\mathbf C$ a category with an initial object named $0$.

Is there a name for the pair of arrows $f,g\colon A\to B$ such that the unique arrow $0\to A$ is their equalizer? And dually, is there a special name for $f,g\colon A\to B$ such that the coequalizer is $B\to 1$, when $1$ is the terminal object of $\mathbf C$? Finally, is it useful to name them? :)

I can figure out how it works in case $\mathbf C$ is concrete: I want to map the fact that a couple of arrows is "everywhere equal" (if "coker(f,g)=the whole") or "nowhere equal" (if "ker(f,g)=nothing unnecessary").

No ideas for general situation + I'm searching references (something make me think about Lawvere but I'm not able to recover anything).

Thanks a lot!

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The former might be called "having disjoint images", at least if C is extensive. I think "everywhere equal" is the wrong intuition for the dual notion; for instance, $id_N\colon N\to N$ and $(+1)\colon N\to N$ have both properties. – Mike Shulman Mar 17 '11 at 18:14
I agree with Mike's comment about the intuition "everywhere equal", and that the former is something to do with disjointness. In fact Diers calls a pair with the latter property "codisjointed". – Steve Lack Mar 18 '11 at 12:19
For the first, surely "disjoint" (or even just "unequal" or "apart") would be enough: it's all about limits and no "image" need ever be formed. I wonder what happens under Stone duality if we apply this to parallel homomorphism that agree only on constants. – Paul Taylor Mar 19 '11 at 18:23
For the dual property, the situation in $\bf Set$ is a binary relation whose equivalence closure is the total relation, or "indiscriminate" as I called it in my book, so maybe we could call this an indiscriminate pair. We could also think of the relation dynamically: it gets you from anywhere to anywhere else, which is called "transitive" in the theory of permutation groups. However, some more examples in other categories would be useful before fixing a name. – Paul Taylor Mar 19 '11 at 18:27
Further to my objection to "disjoint <i>image</i>", consider the identity and swap maps $2\rightrightarrows 2$. These have the property of the question. However, their images are both $2\rightarrowtail 2$ and the intersection of these is again the whole of $2$, not $0$. – Paul Taylor Mar 19 '11 at 19:50