Morphism that is not a mapping I have encountered a statement in Lang's Algebra (Revised Third Edition, page 53) concerning morphisms of objects that seems strange to me: "In practice, in this book we shall see that most of our morphisms are actually mappings, or closely related to mappings."
I have always been under the impression that the terms 'mapping' and 'morphism' are synonymous in the context of categories. Perhaps it is just the case that Lang defines the two in a way that they disagree, but I can't find such an instance. Are they in fact different? If so, what is an example of a morphism that is not a mapping?
 A: For any preordered set $X$, one can define a category whose objects are the elements of $X$, such that $\mathrm{Mor}(x,y)$ has one element if $x\leq y$ and is empty otherwise. This defines a category because $x\leq x$ for all $x\in X$, and if $x\leq y$ and $y\leq z$ then $x\leq z$.
For another example, any monoid $M$ defines a category with one object, whose set of morphisms is the monoid $M$. The composition law is given by the monoid operation.
A: One of my favorite "counterexamples" to preconceived notions about categories is matrix algebra:


*

*The objects are natural numbers

*The arrows are matrices ($\hom(m,n)$ is the collection of $n \times m$ matrices)

*Composition of arrows is the matrix product

A: Here's a naturally-occurring example (coming from computability theory) of a category whose morphisms really aren't "maps" in any good sense: 


*

*The objects are just the sets of natural numbers.

*A pre-morphism from $A$ to $B$ is a Turing machine $\Phi_e$ such that $\Phi_e^A=B$. 
We say that two pre-morphisms $\Phi_{e_0}, \Phi_{e_1}: A\rightarrow B$ are equivalent (and write $e_0\sim e_1$) if for every $X, n$, $$\Phi_{e_0}^X(n)\cong \Phi_{e_1}^X(n)$$ (where "$P\cong Q$" means "either both $P$ and $Q$ are undefined, or $P$ and $Q$ are both defined and are equal to each other").


*

*A morphism is then an equivalence class of pre-morphisms: $$Hom(A, B)=\{\{e_1: e_1\sim e_0\}: \Phi_{e_0}^A=B\}.$$


Now each object is a set . . . but the morphisms don't act as functions between those sets! (This is the fundamental difference between many-one reduction and Turing reduction.) Meanwhile, each morphism is a function - specifically, a partial function from $2^\omega$ to $2^\omega$ - but this functional behavior isn't really reflected in what a given morphism does to a given object! So this category doesn't really satisfy the intuition that morphisms are functions between objects.
