TITLE: Global elements in categories with no terminal object?
QUESTION [16 upvotes]: Let $\mathcal{C}$ be a category.  Suppose $\mathcal{C}$ contains a terminal object, which I will denote by $\boldsymbol{1}$.  Then for any object $B$ in $\mathcal{C}$, a global element of $B$ is a morphism $\boldsymbol{1}\longrightarrow B$. (In many concrete categories, the terminal object is a singleton set, so this definition picks out something close to what we would normally call an "element" of the object $B$.)   Given another object $A$ in $\mathcal{C}$, we can say that a morphism $f:A\longrightarrow B$ is constant if there is some global element $\epsilon:\boldsymbol{1}\longrightarrow B$ such that $f = \epsilon \circ \alpha$, where $\alpha:A\longrightarrow \boldsymbol{1}$ is the (unique) morphism from $A$ to the terminal object $\boldsymbol{1}$.  
So far, these definitions are totally standard and well-known.  But they require the existence of a terminal object.  So my questions are the following:

Suppose $\mathcal{C}$ is a category which does not necessarily have a terminal object.  Is there a suitable definition of "global elements" for the objects of $\mathcal{C}$?  Is there a suitable definition of "constant morphisms"?

Here, by a "suitable" definition, I mean a definition which reduces to the definitions I gave above in the case when $\mathcal{C}$ has a terminal object.  Also a suitable definition of "constant morphism" should have the following property: composing any morphism with a constant morphism yields a constant.  (Note that the definition via terminal objects has this property.)

REPLY [2 votes]: A Paragraph Proof
Say an arrow in a category is “constant” if it's post-composition with any two arbitrary compatible
arrows is indistinguishable.
In the case that there's a terminal object , an arrow $f : a ⟶ b$ is constant precisely when it
factors through the unique arrow $!ₐ : a ⟶ $. This is not terribly difficult to see provided
that “$a$ is non-empty”, i.e., it has a “point”, call it $¡ :  ⟶ a$.
Indeed, if f is constant then the factorization is witness by f ∘ ¡
since $(f ∘ ¡) ∘ !ₐ = f ∘ (¡ ∘ !ₐ) = f ∘ id = f$, where the second equality holds since f is constant. Conversely, if $f$ factorizes through the unique arrow $!ₐ : a ⟶ $ by some arrow $ε$
then, for any composable maps $x,y$ we have $f ∘ x = ε ∘ !ₐ ∘ x = ε ∘ !ₐ ∘ y = f ∘ y$
and so f is constant; where the second equality holds since $!ₐ ∘ x$ and $!ₐ ∘ y$ have the same
type going to  but such arrows are unique.
(Challenge: is the factorisation of constants $a ⟶ b$ via the unique arrow $a ⟶ $ itself unique?)
A Calculational Proof
Below is the originally posted answer, but with the requested notational preliminaries.
• The Booleans are denoted  and have values {true, false}.
• Functions, in type theory or set theory, of the form X →  are called “predicates”.
• Unlike any other equality in mathematics, the equality on  is associative!
  Due to the special statues of this equality, it has a few notations: ≡ and ⇔ being the
  most popular, with = receiving less usage in this context.
  ⟪ I do not use the associativity of boolean equality anywhere in this post;
  an nifty example usage can be found in The associativity of equivalence and the Towers of Hanoi problem
  ⟫
• Rather than write, for example, $∑_{p ∈ ℕ, p \text{ prime}} f(p)$ where the range of the
  quantification is made a second-class citizen in the usual 2-dimensional notation, I use
  linear notation of Z
  and write $(∑ p : ℕ ❙ prime.p • f.p)$.
• More generally, we can define this quantification notation for any monoid $(M, ⊕, u)$
  and write $(⊕ x : X ❙ r.x • f.x)$ for the ⊕ of the terms f.x where x ∈ X satisfies predicate
  $r$.
“Empty range”: (⊕ x : X ❙ false • f.x) = u
“Finite range”: (⊕ i : ℕ ❙ 0 ≤ i ≤ n • f.i) = f.0 ⊕ f.1 ⊕ ⋯ ⊕ f.n
“Abbreviation”: (⊕ x : X • f.x) = (⊕ x : X ❙ true • f.x)

In particular, for the booleans  we have the conjunction ∧, “and”, is associative with unit
  being “true” and so this notation applies to (, ∧, true). It is then conventional to define
  a synonym:  $(∀ x : X ❙ r.x • f.x) ≔ (∧ x : X ❙ r.x • f.x)$. Likewise for the disjunctive monoid
  (, ∨, false) and the ∃xistential quantifier.
• Finally, a proof of an equality can be rendered by a sequence of steps as is done in elementary
  school with the justification of each step annotated between the transitions.
  Informal “P = Q = R where the first equality follows because of reason₁ and the second follows
  because of reason₂” can be rendered in a simpler style as
  P
=⟨ reason₁; i.e, a hint explaining why P = Q ⟩
  Q
=⟨ reason₂; i.e, a hint explaining why Q = R ⟩
  R

The conclusion $P = R$ follows by transitivity of equality.
• This approach to proofs is very popular among Functional Programmers and computer science
  category theorists;
  e.g., A Gentle Introduction to Category Theory --- the calculational approach.
  It is also used in the popular proof-assistant Agda; where it is
  not built-in but is in-fact a user-defined construct! This is possible since Agda allows
  mixfix unicode lexemes as identifiers.
  I am not using any proof assistant for this post.
Below is my original answer.

Let's start with the second question, then follow up with the first.
General definition
Definition 0. for any category , we define the predicate
constant : Arr  → 
constant.f ≡ (∀ x, y • f ∘ x = f ∘ y)

(Where the bullet “•” serves to sepeate the quantifer dummies from the quantifer body).
Definition reduction
We now show that the above definition reduces to that of OP's when terminal objects exist.
Theorem 1. In any category  with terminal object ,
constant.(f : a ⟶ b) ≡ (∃ ε :  ⟶ b • f = ε ∘ !ₐ) , provided ∃ ¡ₐ :  ⟶ a

where !ₓ denotes the unique morphism to the terminal object:
[! characterisation] ∀ x : Obj  • ∀ f : Arr  •     f = !ₓ ≡ f : x ⟶ 

Proof ∷
[⇒] We have $f : a ⟶ b$ and we need to define $ε :  ⟶ b$ and this can be accomplished
if we only had some element $ ⟶ a$. In set theoretical terms, this is tantamount to $a$
being non-empty. ─Challenge: in a category with  and , is it the case that
$x ≠  ≡ ( ⟶ x) ≠ ∅$?─ Anyhow, we can use our proviso here and so define
$ε ≔ f ∘ ¡ₐ$. It remains to show that we have the property OP uses in his/her definition.
   ε ∘ !ₐ
=⟨ definition of ε ⟩ 
   f ∘ ¡ₐ ∘ !ₐ
=⟨ constant.f ⟩
   f ∘ id₁
=⟨ identity ⟩
   f

[⇐] Assuming the existence, let $x, y : p ⟶ a$ then we prove $f ∘ x = f ∘ y$:
   f ∘ x
=⟨ assumption ⟩
   ε ∘ !ₐ ∘ x
=⟨ !-characterisation since !ₐ ∘ x : p ⟶  ⟩
   ε ∘ !ₚ
=⟨ !-characterisation since !ₐ ∘ y : p ⟶  ⟩
   ε ∘ !ₐ ∘ y
=⟨ assumption ⟩  
   f ∘ y 

Constant closure
Finally, we prove the property that OP is interested in, namely:
∀ f,g  • constant.g ⇒ constant.(f ∘ g)

Indeed, given arrows $x$ and $y$, we have
  f ∘ g ∘ x
=⟨ constant.g ⟩
  f ∘ g ∘ y

Notice that this is much simpler than a proof using your more particular definition
─less complexity since we avoid existential quantifiers.
It is interesting to note that
constant.g ≡ (∀ f • constant.(f ∘ g))

Global elements
Let us say that a global element of an object $b$ is any constant map with target $b$, let's denote such elements by a new predicate:
e ⟨∈⟩ b ≡ constant.e ∧ tgt.e = b 

where
f : a ⟶ b ≡ src.f = a ∧ tgt.f = b

Let us show that this reduces to the definition of global elements as we know them when
terminals exist. In particular, let's show that there's a correspondence between the two
notions.
[⇒] given $e ⟨∈⟩ b$, we have some global element $ε :  ⟶ b$ by the reduction theorem
earlier ─of course this relies on us having the same proviso!
[⇐] conversely, given any global element $ε :  ⟶ b$, we know it is constant with target
$b$ and so $ε ⟨∈⟩ b$.
Neato!
Hope this helps :-)