DaoFP Chapter 3 Solutions

solution

Solutions to the exercises of DaoFP, Chapter 3: DaoFP Chapter 3 Exercises. Index: Map of Content.

Solution 3.1.1

Exercise 3.1.1

For $f:a\cong b$, pre-composition $(-\circ f):\mathcal{C}(b,x)\to \mathcal{C}(a,x)$ has inverse $(-\circ f^{-1})$, since $(h\circ f)\circ f^{-1}=h$ and $(k\circ f^{-1})\circ f=k$. See Isomorphism.

Sources: DaoFP Exercise 3.1.1.

Solution 3.1.2

Exercise 3.1.2

${\mathrm{id}}_a$ is its own inverse: ${\mathrm{id}}_a\circ {\mathrm{id}}_a={\mathrm{id}}_a$.

Sources: DaoFP Exercise 3.1.2.

Solution 3.1.3

proof — Exercise 3.1.3

Let $1,1^{\prime }$ be terminal with unique $a:1\to 1^{\prime }$, $b:1^{\prime }\to 1$. Then $b\circ a:1\to 1$ is an arrow into the terminal $1$, hence equals ${\mathrm{id}}_1$; likewise $a\circ b={\mathrm{id}}_{1^{\prime }}$. See Terminal Object.

Sources: DaoFP Exercise 3.1.3.

Solution 3.1.4

proof — Exercise 3.1.4

Any isomorphism $1\to 1^{\prime }$ is an arrow into a terminal object, of which there is exactly one. “Unique up to unique isomorphism.”

Sources: DaoFP Exercise 3.1.4.

Solution 3.2.1

Exercise 3.2.1

Left: $(g\circ h)\circ f^{-1}$; right: $g\circ (h\circ f^{-1})$; equal by associativity — the naturality is automatic here. See Isomorphism, Natural Transformation.

Sources: DaoFP Exercise 3.2.1.

Solution 3.3.1

proof — Exercise 3.3.1

Set $f^{-1}:={\beta }_a({\mathrm{id}}_a):b\to a$. Naturality with $x=a$, $h={\mathrm{id}}_a$, $g:a\to y$ gives ${\beta }_y(g)={\beta }_y(g\circ {\mathrm{id}}_a)=g\circ {\beta }_a({\mathrm{id}}_a)=g\circ f^{-1}$, so ${\beta }_y=(-\circ f^{-1})$ (DaoFP Exercise 3.3.2); with $f:={\beta }_b^{-1}({\mathrm{id}}_b)$ one checks the two are inverse. See Isomorphism, Yoneda Lemma.

Sources: DaoFP Exercise 3.3.1.

Solution 3.3.2

Exercise 3.3.2

${\beta }_y(g)=g\circ f^{-1}$: the whole family is determined by its value on ${\mathrm{id}}_a$.

Sources: DaoFP Exercise 3.3.2.