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+---
+title: Exercises in Category Theory — 1.2
+published: 2016-02-02
+tags: Category Theory
+---
+
+<div class="exercise">
+Let $\ca{Pno}$ be the category of objects $(A, \alpha, a)$ where $A$ is a set, $\alpha : A \to A$ is a unary function, and $a \in A$ is a nominated Element and morphisms $\arr{(A, \alpha, a)}{}{(B, \beta, b)}$ which are functions $f: A \to B$ preserving the structure such that $f \circ \alpha = \beta \circ f$ and $f(a) = b$.
+
+ a) Verify that $\ca{Pno}$ is a category
+ 
+     1. For every $A \in \ca{Pno}$ there exists $\idarr{(A, \alpha, b)}$
+        <div class="proof">
+        $\id$ on $A$ is indeed a function which preserves the structure ($\id \circ \alpha = \alpha = \alpha \circ \id$, $\id(a) = a$) and thus a morphism
+        </div>
+        
+     2. There exists a partial binary operation $\circ$ on the arrows of $\ca{Pno}$
+        <div class="proof">
+        Given three objects $(A, \alpha, a), (B, \beta, b), (C, \gamma, c)$ and two functions $g: A \to B$ and $f: B \to C$ the function $f \circ g: A \to C$ is an arrow in $\ca{Pno}$, that is to say, it preserves structure:
+        $$(f \circ g) \circ \alpha = f \circ \beta \circ g = \gamma \circ (f \circ g)$$
+        and
+        $$(f \circ g)(a) = f(b) = c$$
+        </div>
+ 
+ b) Show that $(\N, \textrm{succ}, 0)$ is a $\ca{Pno}$-object
+ 
+    <div class="proof">
+    $\N$ is indeed a Set, $\textrm{succ}$ is indeed an unary function and $0$ is indeed an element of $\N$.
+    </div>
+    
+ c) Show that for each $\ca{Pno}$-object $(A, \alpha, a)$ there is an unique arrow
+    $$\arr{(\N, \textrm{succ}, 0)}{}{(A, \alpha, a)}$$
+    and describe the behaviour of the carrying function.
+    
+    <div class="proof">
+    We construct a carrying function recursively:
+    $$\begin{aligned}
+    f : \N & \to A \\
+    0 & \mapsto a \\
+    \textrm{succ}(x) & \mapsto \alpha(f(x))
+    \end{aligned}$$
+    $f : \N \to A$ is indeed a morphism and thus an arrow.
+    
+    Given two morphisms $f : \N \to A$ and $g : \N \to A$ we show that they are pointwise identical by induction over $\N$:
+      * $f(0) = a = g(0)$
+      * Given $n \in \N$:
+      $$(f \circ \mathrm{succ})(n) = (\alpha \circ f)(n) \overset{\text{ind.}}{=} (\alpha \circ g)(n) = (g \circ \mathrm{succ})(n)$$
+    </div>
+</div>