# Can we say two morphisms to be isomorphic?

The definition of subobject in Wikipedia is this:

let $$A$$ be an object of some category. Given two monomorphisms

$$u : S\to A$$ and $$v : T\to A$$

with codomain $$A$$, we write $$u\le v$$ if $$u$$ factors through $$v$$—that is, if there exists $$\varphi : S\to T$$ such that $$u = v \circ \varphi$$. The binary relation $$\equiv$$ defined by

$$u\equiv v$$ if and only if $$u\le v$$ and $$v\le u'$$

is an equivalence relation on the monomorphisms with codomain $$A$$, and the corresponding equivalence classes of these monomorphisms are the subobjects of $$A$$. (Equivalently, one can define the equivalence relation by $$u\equiv v$$ if and only if there exists an isomorphism $$\varphi:S\to T$$ with $$u = v \circ \varphi$$.)

This article doesn't assign a name to the equivalence relation $$\equiv$$. Can we call it isomorphism, and say that $$u$$ and $$v$$ are isomorphic if $$u\equiv v$$, or is this confusing, because isomorphism is here the morphism $$\varphi$$, and the adjective isomorphic is reserved for objects, not to morphisms? Or does this $$\equiv$$ relation have some another name?

• You wouldn't necessarily call that equivalence relation "isomorphism" unless it was clear in which category they're isomorphic. Though most people would probably guess you meant in the category Shark mentions below (or suitable subcategory thereof). It doesn't have a special name in any case. – Malice Vidrine Oct 19 '18 at 6:17

Call the category $$\mathscr C$$. Your two monomorphisms are equivalent if $$u=v\circ\varphi$$ where $$\varphi$$ is an isomorphism. One can regard $$u$$ and $$v$$ as elements of the slice category $$\mathscr{C}/A$$. Then $$\varphi$$ is an isomorphism in $$\mathscr C$$ iff it's an isomorphism in $$\mathscr C/A$$. Then $$u$$ and $$v$$ are equivalent iff they are isomorphic as elements of $$\mathscr{C}/A$$.
• And it's worth stating the category the isomorphism lives in, since $\psi\circ u=v\circ\varphi$ gives a different notion of isomorphism of $v$ and $u$, which lives in a different category. – Kevin Carlson Oct 19 '18 at 19:43
• Finally, how can one call the equivalence relation $\equiv$ in the original category? – mma Nov 2 '18 at 6:06