I want to proove
Limit of the identity functor Id ∷ C → C is the initial object
I understand that the limit object would be having morphisms going out to every other object. (Altogether forming the universal cone) So thats one condition satisfied as an initial object.
But theres another condition to be satisfied to be an initial object, which is that there must be only one morphism for each other object.
Being a limit object doesn't mean there's only one cone on it. There could be more natural transformations from apex functor to Id functor as long as there's factorizing morphism going into itself right?
How can I proove it?
Just for the purpose of having the question answered on this site:
Suppose the identity functor $\text{Id}\colon C\to C$ has a limit. In particular this is a cone over $\text{Id}$: an object $L$ in $C$ and an arrow $\pi_X\colon L\to X$ for every object $X$ in $C$, such that for every arrow $f\colon Y\to X$ in $C$, we have $\pi_X = f\circ \pi_Y$.
First we argue that $\pi_L = \text{id}_L$. Note that $\pi_L\colon L\to L$ is an arrow from this cone to itself: since for any object $X$ in $C$, taking $Y = L$ and $f = \pi_X$ in the equation above, we have $\pi_X = \pi_X \circ \pi_L$. But since the cone is terminal, $\text{id}_L$ is the unique arrow from the cone to itself, and $\pi_L = \text{id}_L$.
Next we argue that for any object $X$ in $C$, $\pi_X$ is the unique arrow $L\to X$. Indeed, for any arrow $f\colon L\to X$, taking $Y = L$ in the equation above, we have $\pi_X = f\circ \pi_L = f\circ \text{id}_L = f$.
We conclude that $L$ is an initial object in $C$.