Adjunction from $C$ to $C^{\to}$ and terminal objects In order to practice exercises in unspecified categories I tried the following exercise. Suppose $C$ is a category with a terminal object, then the functor $\mathrm{dom} \colon C^{\to}\to C$ admits a right adjoint.
I was able to do this, with the functor which sends an object $c$ to the unique map from $c$ to the terminal object. This lead me to wonder this was an if and only if. I.e does there exist an adjunction between $C$ and $C^{\to}$, such that the functor out of $C^{\to}$ is the left adjoint, if and only if $C$ has a terminal object. But this it turns out, is too hard for me to be able to have any noteworthy attempt.
Thanks in advance
 A: $\newcommand{\dom}{\text{dom}}$
For clarity, I will restate your claim.

Claim: $\dom : C^\to \to C$ is a left adjoint if, and only if, $C$ has a terminal object.

This does not hold in general. It sounds like you have already shown the if direction is true, and indeed I think it is. I claim the only if direction is false. So I will provide a counter example by showing that there is a category such that $\dom$ is a left adjoint, but $C$ does not have an initial object. Consider any non-trivial groupoid $G$. This groupoid will not have an initial object. Now, we seek to define a right adjoint $F$ to $\dom: G^\to \to G$. It is sufficent to, for each $c\in G$, define a universal arrow $(s , \epsilon_c : \dom (s) \to c)$. See, e.g., Category Theory in Context Lemma 4.6.1 (avalible for free on the authors website). Since $s$ will end up being $F(c)$, I will abuse notation and just write $F(c)$ for $s$.
We set $F(c):\equiv 1_c : c \to c$ and $\epsilon_c:\equiv 1_c$. Now take any arrow out of $\dom$, $((f : x \to y), (h : \dom(f) \to c))$. We need to construct $j$ and $k$ as in the following diagram:

where both the square and the triangle commute. We take $j:\equiv h$ and $k:\equiv h\circ f^{-1}$. Since $G$ is a groupoid, this inverse exists. It is easy to verify everything commutes. Thus, $\dom$ admits a right adjoint, but $G$ does not have an initial object. This gives a large swath of counter examples, since "most" groupoids are non-trivial (only one is, up to equivalence).
