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Exercise 11 - Chapter 3

edited June 2018 in Exercises

Let’s make some definitions, based on the pattern in the previous exercise:

1) What is the category 1? That is, what are its objects and morphisms?

2) What is the category 0?

3) What is the formula for the number of morphisms in n for arbitrary \(n \in \mathbb{N} \) ?

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  • 1.
    edited May 2018

    The Category \(\mathbf{1}\) has one object 0, and one arrow \(id_0)\).

    Comment Source: The Category \\(\mathbf{1}\\) has one object 0, and one arrow \\(id_0)\\).
  • 2.
    edited May 2018

    2) 0 is the category with no objects or morphisms (i.e., both sets are \(\emptyset\).

    3) Claim: There are \(\frac{n(n+1)}{2}\) morphisms in n for all \(n\in\mathbb{N}\). Proof: For every two objects, say \(p>q\), in n there is exactly one one morphism, that obtained from the compositions \(f_{p-1}\circ f_{p-2}\circ\cdots\circ f_{q+1}\circ f_q\). By definition of n, these constitute all of the non-identity morphisms in n. So, there are \(\binom{n}{2}=\frac{n(n-1)}{2}\) non-identity morphisms and \(n\) identity morphisms, for a total of \(\frac{n(n+1)}{2}\) morphisms.

    Comment Source:2) **0** is the category with no objects or morphisms (i.e., both sets are \\(\emptyset\\). 3) Claim: There are \\(\frac{n(n+1)}{2}\\) morphisms in **n** for all \\(n\in\mathbb{N}\\). Proof: For every two objects, say \\(p>q\\), in **n** there is exactly one one morphism, that obtained from the compositions \\(f_{p-1}\circ f_{p-2}\circ\cdots\circ f_{q+1}\circ f_q\\). By definition of **n**, these constitute all of the non-identity morphisms in **n**. So, there are \\(\binom{n}{2}=\frac{n(n-1)}{2}\\) non-identity morphisms and \\(n\\) identity morphisms, for a total of \\(\frac{n(n+1)}{2}\\) morphisms.
  • 3.
    edited May 2018

    3) The recursive function is $$ f(n)= \begin{cases} 0, & \text{if $x \le 0$}.\\\\ f(n-1) + n, & \text{otherwise}. \end{cases} $$ The new element has a morphism to each of its predecessors \(n - 1\) and one to itself. Which is equivalent to $$ f(n) = \frac{n(n+1)}{2} = \frac{[n-1]([n-1]+1)}{2} + n $$

    Comment Source:3) The recursive function is \[ f(n)= \begin{cases} 0, & \text{if $x \le 0$}.\\\\ f(n-1) + n, & \text{otherwise}. \end{cases} \] The new element has a morphism to each of its predecessors \\(n - 1\\) and one to itself. Which is [equivalent to](https://forum.azimuthproject.org/discussion/comment/18665/#Comment_18665) \[ f(n) = \frac{n(n+1)}{2} = \frac{[n-1]([n-1]+1)}{2} + n \]
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