4. 함수 (ii): 합성과 역함수Functions (ii): Composition and Inverses

1. First we prove the if part. Let $(x,y) \in f_1$. Then obviously $(x,y) \in f$ since $f_1 \subset f$. By the definition of the domain, $x \in dom\ f_1$ . Moreover, $dom\ f_1 = B$ since $f_1 : B \to A$ is a function (cf. Theorem 4.8). Therefore, $x \in B$. Conversely, we assume $(x,y) \in f$ and $x \in B$. Since $(x,y) \in f$ and $f = f_1 \cup f_2$, $(x,y) \in f_1$ or $(x,y) \in f_2$. It is sufficient to show that $(x,y) \in f_2$ is impossible. Suppose $(x,y) \in f_2$. Then $x \in dom\ f_2 = C$. Since $C \cap B = \emptyset$, $x \notin B$. But it is contradiction to the assumption that $x \in B$. Thus $(x,y) \notin f_2$.
2. The proof of (2) is analogous to that of (1). We skip the detail.
3. By Theorem 3.77 and Theorem 4.8,
   $$dom\ f = dom\ (f_1 \cup f_2) = (dom\ f_1) \cup (dom\ f_2) = B \cup C.$$

   Similarly,

   $$ran\ f = ran\ (f_1 \cup f_2) = (ran\ f_1) \cup (ran\ f_2) \subset A \cup A = A.$$
4. Let $x \in B$. Then since $f_1 : B \to A$ is a function, $(x, f_1(x)) \in f_1$. Moreover, since $f_1 \subset f$, $(x, f_1(x)) \in f$. Therefore, recalling (4.3), we have $f(x) = f_1(x)$. Similarly, if $x \in C$, then $f(x) = f_2(x)$.
5. Let $(x,y_1) \in f$ and $(x,y_2) \in f$. Then since $f = f_1 \cup f_2$, there exist four cases : (Case 1) $(x,y_1) \in f_1$ and $(x,y_2) \in f_2$, (Case 2) $(x,y_1) \in f_2$ and $(x,y_2) \in f_1$, (Case 3) $(x,y_1) \in f_1$ and $(x,y_2) \in f_1$, (Case 4) $(x,y_1) \in f_2$ and $(x,y_2) \in f_2$. First we prove that (Case 1) and (Case 2) cannot happen. Since $f_1$ is a function from $B$ to $A$, $dom\ f_1 = B$. Similarly, $dom\ f_2 = C$. Moreover, the assumption $B \cap C = \emptyset$ implies that $x \in B$ and $x \in C$ are impossible. Therefore, (Case 1) and (Case 2) are impossible. It only remains to consider (Case 3) and (Case 4). Recall that (Case 3: $(x,y_1) \in f_1$ and $(x,y_2) \in f_1$. Since $f_1 : B \to A$ is a function, $y_1 = y_2$. Similarly, for (Case 4) $(x,y_1) \in f_2$ and $(x,y_2) \in f_2$, we have $y_1 = y_2$ since $f_2 : C \to A$ is a function.

1. Due to Lemma 4.24(3), (5), and Theorem 4.8, $f : B \cup C \to A$ is a function.
2. By (1), $f : B \cup C \to A$ is a function. Thus due to Lemma 4.24(4),
   $$f_1(x) = f(x) \qquad \forall x \in B \tag{4.7}$$

   and

   $$f_2(x) = f(x) \qquad \forall x \in C.$$

   Therefore, by (4.5), we have $f_1 = f_{[B]}$ and $f_2 = f_{[C]}$.

3. By (4.7) and (4.6), $f : B \cup C \to A$ is an extension of $f_1$ to $B \cup C$.
4. The proof of (4) is very analogous to that of (3). We skip the detail.

Due to Theorem 4.8, it is sufficient to show that $g \circ f$ satisfies (F2), $dom\ g \circ f = A$, and $ran\ g \circ f \subset C$. First we prove $ran\ g \circ f \subset C$. By Theorem 3.64(4) and Theorem 4.8,

Next we show $dom\ g \circ f = A$. By Theorem 4.8, $ran\ f \subset B = dom\ g$. Thus by Corollary 3.65 and Theorem 4.8,

It only remains to show that $g \circ f$ satisfies (F2). Let $(x,y_1) \in g \circ f$ and $(x,y_2) \in g \circ f$. Then by the definition of the composition, there exist $z_1$ and $z_2$ such that

and

Since $f : A \to B$ is a function, $(x,z_1) \in f$ and $(x,z_2) \in f$ implies $z_1 = z_2$. Thus since $g : B \to C$ is a function, $(z_1,y_1) \in g$ and $(z_2,y_2) = (z_1,y_2) \in g$ implies $y_1 = y_2$. Therefore $g \circ f$ satisfies (F2).

By (4.2) and the definition of the inverse of the graph $f$,

Due to Theorem 4.8, it is sufficient to show that (1) $dom\ f^{-1} = B$, (2) $ran\ f^{-1} \subset A$, and (3) $f^{-1}$ satisfies (F2).

1. Let $y \in dom\ f^{-1}$. Then by the definition of the domain, there exists a $x$ such that $(y,x) \in f^{-1}$. Moreover, by the definition of the inverse of graph, $(x,y) \in f$. Since $f : A \to B$ is a function, $y \in ran\ f \subset B$ by the definition of the range and Theorem 4.8. Conversely, let $y \in B$. Then since $f : A \to B$ is surjective, there exists a $x \in A$ such that $(x,y) \in f$. Moreover, by the definition of the inverse of graph, $(y,x) \in f^{-1}$. Therefore, by the definition of the domain, $y \in dom\ f^{-1}$.
2. Let $x \in ran\ f^{-1}$. Then by the definition of the range, there exists a $y$ such that $(y,x) \in f^{-1}$. By the definition of the inverse, $(x,y) \in f$. Thus $x \in dom\ f$ by the definition of the domain. Moreover, since $f : A \to B$ is a function, $x \in dom\ f = A$ by Theorem 4.8.
3. Assume that $(x,y_1) \in f^{-1}$ and $(x,y_2) \in f^{-1}$. Then $(y_1,x) \in f$ and $(y_2,x) \in f$. Since $f : A \to B$ is injective, $y_1 = y_2$. Therefore (F2) holds for $f^{-1}$.

Due to Lemma 4.33, $f^{-1} : B \to A$ is a function. Thus it is sufficient to show that the function $f^{-1} : B \to A$ is bijective. Let $x \in A$. Then since $f : A \to B$ is a function, there exists a $y \in B$ such that $(x,y) \in f$. Thus $(y,x) \in f^{-1}$ and it implies that $f^{-1}(y) = x$. Therefore, the function $f^{-1} : B \to A$ is surjective. To prove that the function $f^{-1} : B \to A$ is injective, let $(x_1,y) \in f^{-1}$ and $(x_2,y) \in f^{-1}$. Then $(y,x_1) \in f$ and $(y,x_2) \in f$ by the definition of the inverse. Since $f : A \to B$ is a function, $x_1 = x_2$ by (F2).

Since $f : A \to B$ is invertible, $f^{-1} : B \to A$ is a function. Thus by Theorem 4.8, $dom\ f^{-1} = B$. By Theorem 3.64(2), $dom\ f^{-1} = B$ implies $ran\ f = dom\ f^{-1} = B$. Thus $f : A \to B$ is surjective. To prove that $f : A \to B$ is injective, assume that $(x_1,y) \in f$ and $(x_2,y) \in f$. Then $(y,x_1) \in f^{-1}$ and $(y,x_2) \in f^{-1}$. Since $f^{-1} : B \to A$ is a function, we have $x_1 = x_2$ by (F2).

1. Let $(x,y) \in f^{-1} \circ f$. Then there exists a $z$ such that $(x,z) \in f$ and $(z,y) \in f^{-1}$. Moreover, by the definition of the inverse of graph, $(y,z) \in f$. Thus we have $(x,z) \in f$ and $(y,z) \in f$. Since $f$ is injective due to Corollary 4.36, we obtain
   $$x = y.$$

   Since $f : A \to B$ is a function, $x \in dom\ f = A$. Therefore, by the definition of $I_A$, $(x,y) = (x,x) \in I_A$. Conversely, let $(x,y) \in I_A$. Then by the definition of $I_A$, $x = y$ and $x \in A$. It is sufficient to show that $(x,x) \in f^{-1} \circ f$. Since $f : A \to B$ is a function, there exists a $z$ such that $f(x) = z$, i.e. $(x,z) \in f$. By Lemma 4.31, $x = f^{-1}(z)$. Thus $(z,x) \in f^{-1}$. Therefore, by the definition of the composition of graphs, $(x,x) \in f^{-1} \circ f$.

2. Since the proof of (2) is very similar to (1), we skip the detail.

First we prove that $f$ is injective. Let $(x_1,y) \in f$ and $(x_2,y) \in f$. Then $y \in ran\ f \subset B$. Since $g : B \to A$ is a function, there exists $g(y) \in A$ such that

By the definition of the composition $g \circ f$,

Thus by the assumption that $g \circ f = I_A$, we have

Moreover, by the definition of $I_A$,

for some $x_3, x_4 \in A$. Thus applying Theorem 3.54, we obtain

Next we prove that $f : A \to B$ is surjective. Let $y \in B$. Then by the definition of $I_B$, $(y,y) \in I_B$. Moreover by the assumption that $f \circ g = I_B$, we have $(y,y) \in f \circ g$. Thus by the definition of the composition of graphs, there exists a $z$ such that $(y,z) \in g$ and $(z,y) \in f$. Therefore, $y = f(z)$.

Due to Theorem 4.39 and Corollary 4.36, it only remains to show that $f^{-1} = g$. Moreover, by Theorem 4.13, it is sufficient to show that

Let $y \in B$ and recall that $f^{-1} : B \to A$ is a (bijective) function by Corollary 4.36. Then since both $f^{-1} : B \to A$ and $g : B \to A$ are functions, there exists $x_1 \in A$ and $x_2 \in A$ such that $x_1 = f^{-1}(y)$ and $x_2 = g(y)$. Due to Lemma 4.31, $y = f(x_1)$. By Exercise 4.29, the assumption that $g \circ f = I_A$ with Theorem 4.13, and the definition of the identity function on $A$,

Therefore, since $x_2 = g(y)$ and $g : B \to A$ is a function. $x_1 = x_2$, i.e. $f^{-1}(y) = g(y)$.

Since the if part comes from Exercise 4.40, we only prove the only if part. Assume that $f : A \to B$ is injective and let $C = ran\ f$. Then by Corollary 4.9, $f : A \to C$ is a function. It is injective by the assumption and moreover it is surjective since $C = ran\ f$. Thus $f : A \to C$ is bijective and $f^{-1} : C \to A$ is a bijective function by Corollary 4.36. Moreover, by Corollary 4.43 and Exercise 4.29,

Since $A$ is a nonempty class, there exists a $a \in A$. Consider the constant function $K_a : B - C \to A$ (cf. Example 4.18). In particular, recall that $K_a$ is the graph given by $K_a = \{(x,a) : x \in B - C\}$. Define the graph $g = f^{-1} \cup K_a$. Then by Theorem 4.26(1), $g : B \to A$ is a function. We claim that $g$ is a left inverse of $f : A \to B$. To prove this, it is sufficient to show that

due to Theorem 4.13. Let $x \in A$. Then $f(x) \in ran\ f$ and therefore by Exercise 4.29, the definition of $g : B \to A$ with Lemma 4.24(4), and (4.8),

First recall that $g \circ f : A \to C$ is a function due to Theorem 4.27.

1. Let $(x_1,y) \in g \circ f$ and $(x_2,y) \in g \circ f$. Then by the definition of the composition of graphs, there exist $z_1$ and $z_2$ such that
   $$(x_1,z_1) \in f \quad \wedge \quad (z_1,y) \in g$$

   and

   $$(x_2,z_2) \in f \quad \wedge \quad (z_2,y) \in g$$

   Since $g$ is injective, we obtain $z_1 = z_2$. Thus we have $(x_1,z_1) \in f$ and $(x_2,z_2) = (x_2,z_1) \in f$. Finally, since $f$ is injective, we have $x_1 = x_2$.

2. Let $z \in C$. Then since $g$ is surjective, there exists a $y \in B$ such that $(y,z) \in g$. Moreover, since $f$ is surjective, there exists a $x \in A$ such that $(x,y) \in f$. Finally, by the definition of the composition of graphs,
   $$(x,z) \in g \circ f.$$

   Therefore, $g \circ f$ is surjective (to $C$).

3. This follows immediately from (1) and (2).