# What exactly is a morphism in category theory?

Given a category $C$, an opposite category $C^{op}$ is defined as

• $ob(C^{op})=ob(C)$

• For every $x,y\in ob(C)$, $\text{Hom}_{C^{op}}(x,y)=\text{Hom}_C(y,x)$

One can informally think of this as reversing "arrows" in category $C$.

However, in most abstract sense, morphisms are not arrows. So how does reversing arrows correspond to definition of morphisms in $C^{op}$?

Things in $\text{Hom}_C(y,x)$ have a "source" $y$ and "target" $x$. Now these things become $\text{Hom}_{C^{op}}(x,y)$, but my mind keeps going in this way: surely things in $\text{Hom}_{C^{op}}(x,y)$ would have source $x$ and target $y$, so how does something with swapped sources/targets can be in this set?

Edit: Now I think I see what this is all about. So we don't really care what these morphisms are, but the important fact is that they must satisfy some rules regarding compositions etc. So what we really have is a consistent system of things that follow axioms, and that what we call $\text{Hom}$ sets (in particular they don't have sources or targets or even have to be arrows). In opposite category, we impose exactly same system of "morphisms" but just in opposite direction.

• What is your definition of an "arrow"? Typically in category theory, "arrow" and "morphism" are taken to be synonyms. Jun 6 '17 at 12:33
• "In the most abstract sense, morphisms are not arrows" : to the contrary, morphisms are simply arrows. They don't have to he functions, mappings etc., they can simply be arrows, and so swapping the arrow really isn't a big deal Jun 6 '17 at 12:34
• If $f\in \operatorname{Hom}_{C}(x, y)$, then that means that $f$ is represented by an arrow in the directed graph that makes up $C$, and that arrow goes form $x$ to $y$. At the same time, $f$ is a morphism in $C^{op}$, where it is also represented by an arrow, but that arrow goes form $y$ to $x$. Jun 6 '17 at 12:39
• @Omnomnomnom Yep, I now get it. Now I can say that morphism $f:X\to Y$ and $f:Y\to X$ in $C$ and $C^{op}$ respectively are actually the same thing, but we just declared we are using them in opposite direction Jun 6 '17 at 13:46
• It might be less confusing if you don't use the same symbols for $C$ and its opposite. Say you stick a little superscript op on everything. So instead of $ob(C^{op}) = ob(C)$ you say $ob(C^{op}) = \{ x^{op} : x \in X \}$ where these $x^{op}$ are new things in bijection with the objects of $C$. Then you say the morphisms in $C^{op}$ are $\{f^{op} : f$ a morphism of $C\}$. You define $source(f^{op}) := target(f)^{op}$, $f^{op} \circ g^{op} := (g \circ f)^{op}$, etc. This is isomorphic to the definition you quoted but I find it psychologically easier to digest. Jun 6 '17 at 17:20