# Show a category has an initial object

Consider the category wherein:

• Objects are triples $(X,a,\phi)$, where $X$ is a set, $\;a\;$ is an element of $\;X$, and $\;\phi: X \rightarrow X$ is an endomorphism of $\;X$.

• Morphisms $(X, a, \phi) \rightarrow (Y, b, \psi)$ are functions $f : X \rightarrow Y$ between the underlying sets such that $f(a) = b$ and $f \circ \phi = \psi \circ f$.

Show that this category has an initial object.

So, my understanding of the initial object, in not so sophisticated terms, is an object in $\mathcal{C}$ that has a unique morphism to any given object $X$ in $\mathcal{C}$.

With the above understanding of initial object, I believe this question is asking me to determine the initial object and prove it is the initial object. However, I come up short with how to do this. I believe the correct approach is to investigate how the morphisms act and determine which object fills the condition of being initial. However, I feel my higher level algebra is not as strong as it should be to tackle Category Theory.

I am looking for hints and suggestions on where to go with this as well as possible literature others out there have found useful. Cheers.

• Do you see what the initial object would be if you ignored all those endomorphisms? – Tobias Kildetoft Aug 29 '16 at 8:21
• Isn't $\phi(a)=a$ a required condition for the objects $(X, a,\phi)$? – Crostul Aug 29 '16 at 8:35
• If we ignored all those endomorphisms wouldn't the initial object just be the empty set? – Jeremy Aug 29 '16 at 8:54
• No, the empty set is not an object of this category since we are requiring a distinguished element (I meant to ignore the endomorphisms but not this element). – Tobias Kildetoft Aug 29 '16 at 8:59
• @Crostul Why would it be? There's no reason. – Najib Idrissi Aug 29 '16 at 11:42

Hint: Suppose $(X, a, \phi)$ is an initial object. Then, for any object $(Y, b, \psi)$, the unique morphism $f \colon (X, a, \phi) \to (Y, b, \psi)$ satisfies $$f(a) = b$$ $$f(\phi(a)) = \psi(f(a)) = \psi(b)$$ $$f(\phi^2(a)) = f(\phi(\phi(a))) = \psi(f(\phi(a))) = \psi(\psi(b)) = \psi^2(b)$$ $$f(\phi^3(a)) = f(\phi(\phi^2(a))) = \psi(f(\phi^2(a))) = \psi(\psi^2(b)) = \psi^3(b)$$ $$\vdots$$ $$f(\phi^{n+1}(a)) = f(\phi(\phi^n(a))) = \psi(f(\phi^n(a))) = \psi(\psi^n(b)) = \psi^{n+1}(b)$$