DaoFP Chapter 2 Solutions

solution

Solutions to the exercises of DaoFP, Chapter 2: DaoFP Chapter 2 Exercises. Index: Map of Content.

Solution 2.1.1

proof — Exercise 2.1.1

Both sides send $h$ to $g\circ (f\circ h)=(g\circ f)\circ h$ by associativity. Post-composition is thus a Functor: the covariant Hom Functor $\mathcal{C}(x,-)$.

Sources: DaoFP Exercise 2.1.1.

Solution 2.1.2

Exercise 2.1.2

$(h\circ -)\circ ((g\circ -)\circ (f\circ -))$ and $((h\circ -)\circ (g\circ -))\circ (f\circ -)$ both send $k$ to $h\circ g\circ f\circ k$, by associativity of $\circ$ in the category. See Function Composition.

Sources: DaoFP Exercise 2.1.2.

Solution 2.1.3

proof — Exercise 2.1.3

Applied to $h:c\to x$: the right side gives $(h\circ g)\circ f$, the left $h\circ (g\circ f)$; equal by associativity. Pre-composition is a Contravariant Functor — the contravariant Hom Functor $\mathcal{C}(-,x)$.

Sources: DaoFP Exercise 2.1.3.

Solution 2.3.1

Exercise 2.3.1

Nothing: ${\mathrm{id}}_a\circ h=h$ and $k\circ {\mathrm{id}}_a=k$ by the identity laws. See Identity Function, Category.

Sources: DaoFP Exercise 2.3.1.

Solution 2.4.1

proof — Exercise 2.4.1

Let $x:1\to a$ and $g_1,g_2:c\to 1$ with $x\circ g_1=x\circ g_2$. Since $1$ is terminal, $g_1=g_2={!}_c$ regardless. So $x$ is mono.

Sources: DaoFP Exercise 2.4.1.

Solution 2.5.1

proof — Exercise 2.5.1

Let $!:a\to 1$ and $g_1,g_2:1\to c$ with $g_1\circ !=g_2\circ !$. If $a$ has a Global Element $s:1\to a$ (so $!\circ s={\mathrm{id}}_1$, i.e. $s$ is a section of $!$), then $g_1=g_1\circ !\circ s=g_2\circ !\circ s=g_2$, so $!$ is epi. This is the intended reading (in $\mathbf{Set}$: every nonempty set surjects onto the point). Caveat: for $a=\varnothing$ the map $\varnothing \to 1$ is not epi in $\mathbf{Set}$, since the two maps $1\to \underline{2}$ agree on $\varnothing$ — so the statement needs $a$ to have an element.

Sources: DaoFP Exercise 2.5.1.