Tao Ananlysis I - 3.3.8

Exercise 3.3.8

If $X$ is a subset of $Y$, let $\iota_{X \to Y} : X → Y$ be the inclusion map from $X$ to $Y$, defined by mapping $x \mapsto x$ for all $x∈X$, that is, $\iota_{X→Y}(x):=x$ for all $x∈X$. The map $\iota{X \to X}$ is in particular called the identity map on $X$.

(a) Show that if $X \subseteq Y \subseteq Z$ then $\iota_{Y \to Z} \circ \iota_{X \to Y}=\iota_{X \to Z}$.

(b) Show that if $f: A \to B$ is any function, then $f = f \circ \iota_{A\to A} =\iota_{B \to B} \circ f$.

(c) Show that, if $f:A \to B$ is a bijective function, then $f \circ f^{−1}=\iota_{B \to B}$ and $f^{−1} \circ f= \iota_{A \to A}$.

(d) Show that if $X$ and $Y$ are disjoint sets, and $f : X \to Z$ and $g: Y \to Z$ are functions, then there is a unique function $h: X∪Y \to Z$ such that $h \circ \iota_{X \to X \cup Y} =f$ and $h \circ \iota_{Y\to X \cup Y} =g$.

(e) Show that the hypothesis that $X$ and $Y$ are disjoint can be dropped in (d) if one adds the additional hypothesis that $f(x)=g(x)$ for all $x \in X \cap Y$.

By the way, that symbol $\iota$ is the greek letter iota.

(a) Show that if $X \subseteq Y \subseteq Z$ then $\iota_{Y \to Z} \circ \iota_{X \to Y}=\iota_{X \to Z}$

Let's first explain the meaning of $X \subseteq Y \subseteq Z$. It says that every $x \in X$ is also in $Y$, and that every $y \in Y$ is also in $Z$, resulting in every $x \in X$ also being in $Z$.

Now, let's consider $\iota_{X \to Y}$. This is a function from $X$ to $Y$, mapping $x \mapsto x$.

Similarly, $\iota_{Y \to Z}$ is a function from $Y$ to $Z$, mapping $y \mapsto y$.

So, the composition $\iota_{Y \to Z} \circ \iota_{X \to Y}$ is a function $X \to Y \to Z$, mapping $x \mapsto x$, then $y \mapsto y$. That second mapping is equivalent to $x \mapsto x$ because the first one maps $x \mapsto x$, where $x \in Y$.

Thus, $\iota_{Y \to Z} \circ \iota_{X \to Y}$ is a function $X \to Z$, mapping $x \mapsto x$. This is the definition of $\iota_{X \to Z}$. $\square$

(b) Show that if $f: A \to B$ is any function, then $f = f \circ \iota_{A\to A} =\iota_{B \to B} \circ f$.

Let's first consider $\iota_{A to A}$.  This is a function $A \to A$ mapping $a \mapsto a$, where $a \in A$.

The composition $f \circ \iota_{A to A}$ is therefore a function $A \to A \to B$, mapping $a \mapsto a$ then $a \mapsto f(a)$, where $f(a) \in B$. This is equivalent to a function $A \to B$, mapping $a \mapsto f(a)$. This is the definition of $f$. 

So we have shown $f = f \circ \iota_{A\to A}$. $\square$

Let's now consider the function $\iota_{B \to B}$. This is a function $B \to B$ which maps $b \mapsto b$, for $b \in B$. 

The composition $\iota_{B \to B} \circ f$ is a function $A \to B \to B$, which first maps $a \mapsto f(a)$ then $b \mapsto b$. Since $f(a) \in B$, the second mapping is $f(a) \mapsto f(a)$. The composition is therefore a function $A \to B$, mapping $a \mapsto f(a)$. This is the definition of $f$.

So we have shown $f = \iota_{B \to B} \circ f$. $\square$

(c) Show that, if $f:A \to B$ is a bijective function, then $f \circ f^{−1}=\iota_{B \to B}$ and $f^{−1} \circ f= \iota_{A \to A}$.

Let's first consider $f \circ f^{-1}$. This composition is a function $B \to A \to B$. Because $f$ is bijective, it is invertible, and for every $b \in B$ there is a unique $a \in A$ such that $f(a)=b$. That $a$ is $f^{-1}(b)$. 

And so we have $f \circ f^{-1}(b) = f(a) = b$.

The composition is equivalent to a function $B \to B$ which maps $b \mapsto b$. This is the definition of $\iota_{B \to B}$. $\square$

The argument for the second equivalence is very similar.

Consider $f^{-1} \circ f$. This composition is a function $A \to B \to A$. Because $f$ is bijective, it is invertible, and for every $b \in B$ there is a unique $a \in A$ such that $f(a)=b$. That $a$ is $f^{-1}(b)$.  

And so we have $f^{-1} \circ f(a) = f^{-1}(b) = a$.

The composition is equivalent to a function $A \to A$ which maps $a \mapsto a$. This is the definition of $\iota_{A \to A}$. $\square$

(d) Show that if $X$ and $Y$ are disjoint sets, and $f : X \to Z$ and $g: Y \to Z$ are functions, then there is a unique function $h: X∪Y \to Z$ such that $h \circ \iota_{X \to X \cup Y} =f$ and $h \circ \iota_{Y\to X \cup Y} =g$.

Our aim is to show that $h$ is indeed a function, and that it is unique. Let's draw a picture to aid our thinking.

For $h$ to be a function, it needs to have a defined domain, codomain, and map elements from the domain to a single element of the codomain. The domain is $X \cup Y$, and the codomain is $Z$. 

Does $h$ map elements from the domain to a single element of the codomain? Well, an element of the domain is either in $X$ or in $Y$ but not both since $X$ and $Y$ are disjoint. 

If an element is in $X$ then $\iota_{X \to X \cup Y}$ maps $x \mapsto x$ where $x \in X$, but expands the domain $X$  to $X \cup Y$ as the codomain. Importantly, if the input is a particular $x$, the output is the same $x$. So the composition $h \circ \iota_{X \to X \cup Y}$ maps $x \in X$ to $z \in Z$, and we are given this is $f(x)$. Since $f$ is a function, this means $h$ maps each element of $X$ to a single element of $Z$. The same argument applies for $Y$ and $h \circ \iota_{X \to X \cup Y}$, which tells us that $h$ maps each element of $Y$ to a single element of $Z$.

Because $X$ and $Y$ are disjoint, there is no element of $X \cup Y$ which could map to two different elements of $Z$, via $f$ and $g$.

So $h$ has a defined domain, codomain, and we have shown it maps elements of $X \cup Y$ to a single element of $Z$. This means $h$ is a function. $\square$

Let's now show it is unique. Let's assume $h: X∪Y \to Z$ and $h': X∪Y \to Z$ are different functions, subject to the same constraints:

• $h \circ \iota_{X \to X \cup Y} = f$ and $h' \circ \iota_{X \to X \cup Y} = f$
• $h \circ \iota_{Y\to X \cup Y} = g$ and $h' \circ \iota_{Y\to X \cup Y} = g$

The domain of $h'$ is $X \cup Y$. An element of this domain is either in $X$ or $Y$ but not both. Let's consider both cases:

• For $x \in X$, we have $h'( \iota_{X \to X \cup Y}(x))=h'(x)=f(x)$. We also have $h( \iota_{X \to X \cup Y}(x))=h(x)=f(x)$. So $h(x) = h'(x)$ for $x \in X$.
• For $y \in Y$, we have $h'( \iota_{Y \to X \cup Y}(y))=h'(y)=g(y)$. We also have $h( \iota_{Y \to X \cup Y}(y))=h(y)=g(y)$. So $h(y) = h'(y)$ for $y \in Y$.

So $h=h'$ over the entire domain $X \cup Y$. We have shown that $h$ is unique. $\square$

(e) Show that the hypothesis that $X$ and $Y$ are disjoint can be dropped in (d) if one adds the additional hypothesis that $f(x)=g(x)$ for all $x \in X \cap Y$.

Let's draw a picture to help our thinking.

If we drop the requirement for $X$ and $Y$ to be disjoint, we need another way to ensure that $h$ is a function that maps each element of $X \cup Y$ to a single element of $Z$.

We can see that elements in $X \cap Y$ can be mapped by both $f$ and $g$. To ensure the mapping is unique, we need $f(x)=g(x)$ for all $x \in X \cup Y$. $\square$