If $C(c,-)$ maps all morphisms to isomorphisms, what can we say about $c$? Can anyone provide examples, insight, or literature on objects $c$ in a category $C$ where $C(c,-)$ maps all morphisms to isomorphisms?
Another way to say the same thing is that $c$ is left orthogonal to every morphisms in $C$, i.e. for any $v \colon c \to b$ and $f\colon a \to b$ there exists a unique lift $u \colon c \to a$ with $fu = v$.
The only simple examples I can think of are objects which are initial, or merely initial in their connected component. A more complicated example, which prompted the question, is that for any functor $T : A \to B$, a $T$-generic morphism with domain $d$ is precisely an object with this property in the category $d \downarrow T$.
 A: What you are looking for are local objects. I have once searched for reference about them and quickly gave up as I have found nothing usable, only the nLab page. Therefore I will rather provide the insight and examples you have also asked for:
Let $\mathcal{C}$ be a category and $J\subset\operatorname{Ar}\mathcal{C}$ be a class of morphisms, then:
Definition ($J$-local object): An object $c\in\operatorname{Ob}\mathcal{C}$, for which $\operatorname{Hom}_\mathcal{C}(-,c)\colon\mathcal{C}^\mathrm{op}\rightarrow\mathbf{Set}$ maps all morphisms in $J$ to isomorphisms (bijections), is a $J$-local object.
Definition ($J$-local morphism): A morphism $f\in\operatorname{Ar}\mathcal{C}$, for which $\operatorname{Hom}_\mathcal{C}(f,c)$ is an isomorphism (bijection) for all $J$-local objects $c\in\operatorname{Ob}\mathcal{C}$, is a $J$-local morphism.
To give a few results: Immediatly from the definitions, we have:
Corollary: Every morphism in $J$ is a $J$-local morphism.
Corollary: Every isomorphism is a $J$-local morphism.
Since limits preserve isomorphisms, we have:
Lemma: Limits preserve $J$-local objects and colimits preserve $J$-local morphisms.
Since bijections fulfill the two-out-of-three property, we have:
Lemma: $J$-local morphism fulfill the two-out-of-three property.
Corollary: Together with the class of $J$-local morphisms, $\mathcal{C}$ is a category with weak equivalences.
A: I don’t have a (good) answer to the question, but only a slight observation.
Recall.
Given any object $c$ in a category $C$, the set $G ≔ C(c, c)$ becomes a monoid via composition of endomorphisms.
For every object $d$ of $C$, the set $C(c, d)$ is then a right $G$-set via precomposition.
If $f \colon d \to e$ is a morphism in $C$, then the induced map $f_* \colon C(c, d) \to C(c, e)$ is a homomorphism of right $G$-sets.

Let us now consider the given situation.
For every element $g$ of $G$, the left-multiplication map $g_* \colon G \to G$ is bijective.
Therefore:

*

*$G = C(c, c)$ is a group.

If there exists a morphism between two objects $d$ and $e$ of $C$, then $C(c, d)$ in non-empty if and only if $C(c, e)$ is non-empty.
For every object $d$ in the connected component of $c$, the set $C(c, d)$ is therefore non-empty.
In other words, there exists an element $f$ in $C(c, d)$.
The map $f_* \colon G \to C(c, d)$ is an isomorphism of right $G$-sets.
Therefore:


*For every object $d$ in the connected component of $c$, we have $C(c, d) ≅ G$ is right $G$-sets.


The combination of conditions 1 and 2 is in fact equivalent to $C(c, -)$ mapping morphisms to isomorphisms.
Indeed, suppose conversely that the object $c$ satisfies conditions 1 and 2.
Let $f \colon d \to e$ be an arbitrary morphism in $C$.
We have two cases to consider:

*

*If both $d$ and $e$ are outside the connected component of $c$, then both $C(c, d)$ and $C(c, e)$ are empty.
Consequently, $f_* \colon C(c, d) \to C(c, e)$ is bijective.


*If both $d$ and $e$ are contained in the connected component of $c$, then by condition 1, both $C(c, d)$ and $C(c, e)$ are isomorphic to $G$ as right $G$-sets.
The composite
$$
  G ≅ C(c, d) \xrightarrow{\enspace f^* \enspace} C(c, e) ≅ G
$$
is a homomorphism of right $G$-sets.
But every homomorphism of right $G$-sets $G \to G$ is given by left-multiplication with some element of $G$, and therefore bijective by condition 2.
Consequently, $f_* \colon C(c, d) \to C(c, e)$ was bijective to begin with.
