## Chapter - 5, Functions

### Section - 5.3 - Inverses of Functions

### Summary

- Suppose $f : A \to B$. If $f$ is one-to-one and onto, then $f^{-1}: B \to A$.
- Suppose $f$ is a function from $A$ to $B$, and suppose that $f^{−1}$ is a function from $B$ to $A$. Then $f^{−1} \circ f = i_A$ and $f \circ f^{−1} = i_B$.
- Suppose $f : A \to B$. Then,
- If there is a function $g : B \to A$ such that $g \circ f = i_A$ then $f$ is one-to-one.
- If there is a function $g : B \to A$ such that $f \circ g = i_B$ then $f$ is onto.

- Suppose $f : A \to B$. Then the following statements are equivalent.
- $f$ is one-to-one and onto.
- $f^{−1}: B \to A$.
- There is a function $g : B \to A$ such that $g \circ f = i_A$ and $f \circ g = i_B$.

- Suppose $f: A \to B$, $g: B \to A$, $g \circ f = i_A$, and $f \circ g = i_B$. Then $g = f^{−1}$.

**Soln1**

Required function $f^{-1} = \{(p,q) \in P \times P \, \vert \, \text{the person } p \text{ is sitting immediately to the left of the person q} \}$.

**Soln2**

Required function $F^{-1}(X) = A/X$.

**Soln3**

Suppose $g(x) = \frac {3x-5} 2$. Thus we have:

$\Rightarrow g \circ f (x) = g(f(x)) = \frac {3 \times {\frac {2x+5} 3 } - 5} 2$.

$\Rightarrow g \circ f (x) = \frac {2x+5 - 5} 2$.

$\Rightarrow g \circ f (x) = \frac {2x} 2$.

$\Rightarrow g \circ f (x) = x$.

Thus we have $g \circ f = i_{\mathbb R}$.

Also, we have $f \circ g(x) = \frac {2 \times {\frac {3x - 5} 2} + 5} 3$.

$\Rightarrow f \circ g(x) = \frac {3x - 5 + 5} 3$.

$\Rightarrow f \circ g(x) = \frac {3x} 3$.

$\Rightarrow f \circ g(x) = x$.

Thus we have $f \circ g = i_{\mathbb R}$.

Since $f \circ g = i_{\mathbb R}$ and $g \circ f = i_{\mathbb R}$, it follows that $f$ is one-to-one and onto and $g = f^{-1}$.

**Soln4**

Suppose $g(x) = {\frac {x+3} 2}^{\frac 1 3}$. Clearly $g(x)$ is defined for all values of $x \in \mathbb R$(cube root of negative number is defined). Thus $g: \mathbb R \to \mathbb R$.

Now, $g \circ f(x) = {(\frac {(2x^3 - 3) +3} 2)}^{\frac 1 3}$.

$\Rightarrow g \circ f(x) = {(\frac {2x^3} 2)}^{\frac 1 3}$.

$\Rightarrow g \circ f(x) = {(x^3)}^{\frac 1 3}$.

$\Rightarrow g \circ f(x) = x$.

Thus we have $g \circ f = i_{\mathbb R}$.

Also, $f \circ g(x) = 2({(\frac {x + 3} 2)^{\frac 1 3}})^3 - 3$.

$\Rightarrow f \circ g(x) = 2{\frac {x + 3} 2} - 3$.

$\Rightarrow f \circ g(x) = x + 3 - 3$.

$\Rightarrow f \circ g(x) = x$.

Thus we have $f \circ g = i_{\mathbb R}$.

Since $f \circ g = i_{\mathbb R}$ and $g \circ f = i_{\mathbb R}$, it follows that $f$ is one-to-one and onto and $g = f^{-1}$.

**Soln5**

Suppose $g(x) = 2 - \log x$. Clearly $g: R^+ \to R$.

We have $g \circ f(x) = 2 - \log 10^{2-x} = 2 - (2 - x) = x$. Thus $g \circ f = i_{\mathbb R}$.

Also, we have $f \circ g(x) = 10^{2 - (2 - \log x)} = 10^{\log x} = x$. Thus $f \circ g = i_{\mathbb R^+}$.

Since $f \circ g = i_{\mathbb R^+}$ and $g \circ f = i_{\mathbb R}$, it follows that $f$ is one-to-one and onto and $g = f^{-1}$.

**Soln6**

Suppose $g(x) = \frac {2x} {x - 3}$ is a function from $B = \mathbb R \setminus \{ 3 \}$ to $\mathbb R$. Clearly $g(x) \ne 2$ for any value of $x$. Thus we can say that $g: B \to A$, where $B = \mathbb R \setminus \{ 3 \}$ and $A = \mathbb R \setminus \{ 2 \}$.

Now, $g \circ f(x) = \frac {2{\frac {3x} {x-2} } } { { \frac {3x} {x-2} } - 3 }$.

$\Rightarrow g \circ f(x) = \frac {\frac {6x} {x-2} } { { \frac {3x - 3(x-2)} {x-2} } }$.

$\Rightarrow g \circ f(x) = \frac {\frac {6x} {x-2} } { { \frac {6} {x-2} } }$.

$\Rightarrow g \circ f(x) = \frac {6x} 6$.

$\Rightarrow g \circ f(x) = x$.

Thus we have $g \circ f = i_A$.

Similarly it can be shown that, $f \circ g(x) = i_B$.

Since $f \circ g = i_B$ and $g \circ f = i_A$, it follows that $f$ is one-to-one and onto and $g = f^{-1}$.

Thus **(a)** and **(b)** both are solved.

**Soln7**

**(a)**

To prove $f = f_2 \circ f_1$, we need to prove that $\forall x \in \mathbb R (f(x) = f_2 \circ f_1 (x))$. Suppose $x \in \mathbb R$. Thus $f_2 \circ f_1(x) = f_2 (f_1(x)) = \frac {f_1(x)} 5 = \frac {x+7} 5 = f(x)$. Since $x$ is arbitrary, $f = f_2 \circ f_1$.

**(b)**

Clearly ${f_1}^{-1}(x) = x - 7$ and ${f_2}^{-1}(x) = 5x$. Thus ${f_1}^{-1} \circ {f_2}^{-1}(x) = {f_2}^{-1}(x) - 7 = 5x - 7 = f^{-1}(x)$.

**Soln8**

**(a)**

Suppose $b$ is an arbitrary element of $B$. Let $a = f^{-1}(b) \in A$. Then $(b,a) \in f^{-1}$, so $(a, b) \in f$ and therefore $f(a) = b$.

Thus, $(f \circ f^{-1})(b) = f( f^{-1} (b)) = f(a) = b = i_B(b)$.

Since $b$ was arbitrary, we have shown that $\forall b \in B(( f^{−1} \circ f)(b) = i_B(b))$, so $f \circ f^{−1} = i_B$.

**(b)**

Suppose $g = f^{-1}$. Since $f = (f^{-1})^{-1}$, it follows $f \circ f^{-1} = (f^{-1})^{-1} \circ f^{-1} = g^{-1} \circ g$. Since $g = f^{-1}$, $g: B \to A$. Thus $g^{-1} \circ g = i_B$(using the first half of the proof as required in the solution). It follows $f \circ f^{-1} = i_B$.

**Soln9**

Suppose $f: A \to B$. Suppose $g: B \to A$ such that $f \circ g = i_B$. Suppose $b \in B$. Thus $(b,b) \in i_B$. It follows that $(b,b) \in f \circ g$. Thus There exist an element $a \in A$ such that $(b,a) \in g$ and $(a,b) \in f$. Thus there exist an element $a \in A$ such that $f(a) = b$. Since $b$ is arbitrary, it follows that $\forall b \in B \exists a \in A ( f(a) = b )$. Thus $f$ is onto.

**Soln10**

($\to$)Suppose $(b,a) \in B \times A$ is an arbitrary pair in $g$. Since $b \in B$, it follows $(b,b) \in i_B$. Since $f \circ g = i_B$, it follows $(b,b) \in f \circ g$. Thus for some element $a' \in A$, $(b,a') \in g$ and $(a',b) \in f$. Since $g$ is a function and $(b,a) \in g$, it follows $a = a'$. Thus $(a,b) \in f$. It follows $(b,a) \in f^{-1}$. Thus $g \subseteq f^{-1}$.

($\leftarrow$) Suppose $(b,a) \in B \times A$ is an arbitrary pair in $f^{-1}$. Thus $(a,b) \in f$ and $(a,a) \in i_A$. Since $g \circ f = i_A$, it follows $(a,a) \in g \circ f$. Thus for some element $b' \in B$, $(a,b') \in f$ and $(b',a) \in g$. Since $f$ is a function and $(a,b) \in f$, it follows $b = b'$. Thus $(b,a) \in g$. Thus $f^{-1} \subseteq g$.

Since $f^{-1} \subseteq g$ and $g \subseteq f^{-1}$, it follows that $g = f^{-1}$.

**Soln11**

**(a)**

Since $f: A \to B$ and $g: B \to A$ and $f \circ g = i_B$, it follows that $f$ is onto by the theorem(3.1 in above summary). It is given that $f$ is one-to-one. Thus $f$ is one-to-one and onto. Thus $f^{-1} \circ f = i_A$.

Thus, $g = i_A \circ g = (f^{-1} \circ f) \circ g = f^{-1} \circ (f \circ g) = f^{-1} \circ i_B = f^{-1}$.

**(b)**

Since $f: A \to B$ and $g: B \to A$ and $g \circ f = i_A$, it follows that $f$ is one-to-one by the theorem(3.2 in above summary). It is given that $f$ is onto. Thus $f$ is one-to-one and onto. Thus $f \circ f^{-1} = i_B$.

Thus, $g = g \circ i_B = g \circ (f \circ f^{-1})= (g \circ f) \circ f^{-1} = i_A \circ f^{-1} = f^{-1}$.

**(c)**

Since $f: A \to B$ and $g: B \to A$ and $f \circ g = i_B$, it follows from theorem(3rd theorem form above summary) that $f$ is onto and $g$ is one-to-one.

Proof for $g$ is not onto:

Suppose $g$ is onto. Thus $g$ is one-to-one and onto. Thus by theorem, $g^{-1} = f$. Applying $g$ on both sides, $g \circ g^{-1} = g \circ f$. Since $g$ is one-to-one and onto, $g \circ g^{-1} = i_A$. Thus $g \circ f = i_A$. But it contradicts with the given that $g \circ f \ne i_A$. Thus we can conclude that $g$ is not onto.

Proof for $f$ is not one-to-one:

Suppose $f$ is one-to-one. Thus $f$ is one-to-one and onto. Thus by theorem, $f^{-1} = g$. Thus $f^{-1} \circ f = g \circ f$. Since $f$ is one-to-one and onto, $f^{-1} \circ f = i_A$. Thus $g \circ f = i_A$. But it contradicts with the given that $g \circ f \ne i_A$. Thus we can conclude that $f$ is not one-to-one.

**Soln12**

Suppose $f: A \to B$ is one-to-one. Suppose $B' = Ran(f)$. Clearly $B' \subseteq B$. Now we shall prove $f^{-1}: B' \to A$. Thus we shall do it by proving existence and uniqueness:

*Existence:*

Suppose $b \in B'$. Thus $b \in Ran(f)$. It follows that there exists $a \in A$ such that $(a,b) \in f$. Thus $(b,a) \in f^{-1}$. Thus there exist $a \in A$ such that $(b,a) \in f^{-1}$. Since $b$ is arbitrary, it follows $\forall b \in B' \exists a \in A ((b,a) \in f^{-1})$.

*Uniqueness:*

Suppose $b \in B'$. Suppose $(b,a_1) \in f^{-1}, (b,a_2) \in f^{-1}$. Thus $(a_1,b) \in f$ and $(a_2,b) \in f$. Since $f$ is one-to-one, it follows $a_1 = a_2$. Thus there exists a unique element $a \in A$ such that $(b,a) \in f^{-1}$.

Thus $f^{-1}: B' \to A$.

**Soln13**

**(a)**

We have $R = \{ (x,y) \in A \times A \, \vert \, f(x) = f(y) \}$. It follows that $\forall x \in A \forall y \in A (xRy \leftrightarrow f(x) = f(y))$. Clearly we have $f$ is compatible with $R$, since $\forall x \in A \forall y \in A(xRy \to f(x) = f(y))$. Thus it follows directly from Ex18 from sec-5.1, that $h = \{ (X,y) \in A/R \times B \, \vert \, \exists x \in X (f(x) = y) \}$ is a function such that $h([x]_R) = f(x)$.

**(b)**

Since $\forall x \in A \forall y \in A (xRy \leftrightarrow f(x) = f(y))$, it follows from ex16 from sec5.2 that $h$ is one-to-one.

Proof for $h$ is onto. Suppose $b \in B$. Since $f$ is onto, it follows that there exists an $a \in A$ such that $f(a) = b$. Since $a \in A$ and $R$ is an equivalence relation, it follows $[a]_R \in A/R$. Thus $h([a]_R) = f(a) = b$. Since $b$ is arbitrary, it follows that for every $b \in B$, there exists an element $[a]_R \in A/R$. Thus $h$ is onto.

**(c)**

Since $h: A/R \to B$ is one-to-one and onto, it follows that $h^{-1}: B \to A/R$ exists. Suppose $(b,X) \in h^{-1}$. Thus $(X,b) \in h$. It follows there exist $x \in A$ such that $[x]_R = X$ and $b = f(x)$. Thus $([x]_R,b) \in h$. Suppose $y \in [x]_R$. Thus $yRx$, or $f(y) = f(x)$. Since $f(x) = b$, it follows $f(y) = b$. Thus $\forall y \in [x]_R (f(y) = b)$. Since $X = [x]_R$, it follows $\forall y \in X (f(y) = b)$. Since $(b,X) \in h^{-1}$, it follows $h^{-1}(b) = \{ y \in A \, \vert \, f(y) = b \}$.

**(d)**

It seems like there is a little typo in the question, instead of $\forall b \in B(g(b) \in h(b))$, it seems it actually should be $\forall b \in B(g(b) \in h^{-1}(b))$.

Suppose $g: B \to A$.

($\to$) Suppose $f \circ g = i_B$. Suppose $b \in B$. Thus there must exist an element $a \in A$ such that $(b,a) \in g$, or $g(b) = a$. Thus $f(g(b)) = f(a)$, which is equivalent to $f \circ g(b) = f(a)$. Since $f \circ g = i_B$, it follows $b = f(a)$, or $f(a) = b$. Thus from part(c) of this solution it follows that $a \in h^{-1}(b)$. Since $a = g(b)$, it follows $g(b) \in h^{-1}(b)$. Since $b$ is arbitrary, it follows $\forall b \in B (g(b) \in h^{-1}(b))$.

($\leftarrow$) Suppose $\forall b \in B (g(b) \in h^{-1}(b))$. Suppose $b \in B$ is an arbitrary element. Thus there must exist an element $a \in A$ such that $(b,a) \in g$, or $g(b) = a$. Thus $a \in h^{-1}(b)$. By part(c) it follows that $f(a) = b$. Since $a = g(b)$, it follows $f(g(b)) = b$. Thus $f \circ g(b) = b$. Since $b$ is arbitrary, it follows that $f \circ g = i_B$.

**Soln14**

**(a)**

Suppose $x \in A'$. Thus $x \in Ran(g)$. It follows that there exist an element $b \in B$ such that $g(b) = x$. Thus we have $g \circ f(x) = g(f(x)) = g(f(g(b))) = g(f \circ g(b)) = g(i_B(b)) = g(b) = x$. Since $x$ is arbitrary, it follows that $g \circ f(x) = x$ for all $x \in A'$.

**(b)**

We know from Sec5.1 Ex-7, $f \upharpoonright A' = f$. Since $f \circ g = i_B$, it follows $(f \upharpoonright A') \circ g = i_B$. Also from part(a) of this solution we have $g \circ (f \upharpoonright A') = i_{A'}$. Thus $f \upharpoonright A'$ is one-to-one and onto and $(f \upharpoonright A')^{-1} = g$.

**Soln15**

Suppose $A' = Ran(g)$. Thus clearly $A' = B$. Now with $A' = Ran(g)$, it follows from Soln14(b) that $(f \upharpoonright A')^{-1} = g$. Since $A' = B$, it follows $(f \upharpoonright B)^{-1} = g$.

**Soln16**

**(a)**

We have $f(x) = 4x - x^2$. Suppose $y = f(x)$. Thus we have $4x - x^2 = y$. Thus $x^2 -4x +y = 0$. It is a quadratic equation, thus we have $x = 2 \pm \sqrt {4-y}$. Clearly $y \le 4$. Thus $B = \{ y \in \mathbb R \, \vert \, y \le 4 \}$.

**(b)**

Suppose $g: B \to A$ such that $g(x) = 2 + \sqrt {4-x}$ where $B = \{ x \in \mathbb R \, \vert \, x \le 4 \}$ and $A = \{x \in \mathbb R \, \vert \, x \ge 2 \}$.
Thus we have $g \circ f(x) = g(f(x)) = 2 + \sqrt {4-f(x)} = 2 + \sqrt {4 - (4x - x^2)} = 2 + \sqrt {4 - 4x + x^2)} = 2 + \sqrt {(x-2)^2} = 2 + x - 2 = x$.
Thus $g \circ f(x) = i_B$.

Similarly $(f \upharpoonright A) \circ g(x) = 4g(x) - (g(x))^2 = g(x)(4 - g(x))$. Thus $(f \upharpoonright A) \circ g(x) = (2 + \sqrt {4-x})(4 - 2 - \sqrt {4-x})) = (2 + \sqrt {4-x})(2 - \sqrt {4-x}) = 2^2 - (\sqrt {4-x})^2 = x$.
Thus $(f \upharpoonright A) \circ g(x) = i_A$. Now from Soln14(b) we have $(f \upharpoonright A)^{-1} = g$.

**Soln17**

We shall prove $f = g^{-1}$ by proving that $f \circ g = i_{\mathcal S}$ and $g \circ f = i_{\mathcal P}$, where $\mathcal P$ is set of all partial orders and $\mathcal S$ is set of all strict partial order.

Proof of $f \circ g = i_{\mathcal S}$:

Suppose $R \in \mathcal S$. Thus $R$ is strict partial order. Thus $R$ and $i_{\mathcal A}$ are disjoint. It follows $f \circ g(R) = f(g(R)) = (R \cup i_A) \setminus i_A = R \setminus i_A$. Since $R$ and $i_A$ are disjoint, it follows $f \circ g (R) = R \setminus i_A = R$. Thus $f \circ g = i_{\mathcal S}$.

Proof of $g \circ f = i_{\mathcal P}$:

Suppose $R \in \mathcal P$. Thus $R$ is partial order. Thus $i_{\mathcal A} \subseteq R$. It follows $g \circ f(R) = g(f(R)) = (R \setminus i_A) \cup i_A = R \cup i_A$. Since $i_A \subseteq R$, it follows $g \circ f (R) = R \cup i_A = R$. Thus $g \circ f = i_{\mathcal P}$.

Thus $f$ is one-to-one and onto and $g = f^{-1}$.

**Soln18**

**(a)**

To prove $R$ is an equivalence relation, we need to prove that $R$ is reflexive, symmetric and transitive.

__Reflexive:__

Suppose $f \in \mathcal F$. Suppose $h = i_A$. Thus $f = i_A \circ f \i_A$, or $f = (i_A)^{-1} \circ f \circ i_A$. It follows $fRf$. Thus $R$ is reflexive.

__Symmetric:__

Suppose $f,g \in \mathcal F$ such that $fRg$. Thus there exist an $h \in \mathcal P$ such that $f = h^{-1} \circ g \circ h$. It follows $h \circ f = g \circ h$(by applying h on both sides). Similarly taking value corresponding to $h^{-1}$ on both sides, $h \circ f \circ h^{-1} = g \circ h \circ h^{-1} = g$. Thus $gRf$. Thus $R$ is symmetric.

__Transitive:__

Suppose $f,g,h \in \mathcal F$ such that $fRg$ and $gRh$. Thus for some $h_1, h_2 \in \mathcal P$, we have $f = h_1^{-1} \circ g \circ h_1$ and $g = h_2^{-1} \circ h \circ h_2$. Thus we have $f = h_1^{-1} \circ (h_2^{-1} \circ h \circ h_2) \circ h_1 = (h_1^{-1} \circ h_2^{-1}) \circ h \circ (h_2 \circ h_1) = (h_2 \circ h_1)^{-1} \circ h \circ (h_2 \circ h_1)$. Thus $fRh$. It follows $R$ is transitive.

Thus $R$ is an equivalence relation.

**(b)**

Suppose $fRg$. Thus there exist an $h \in \mathcal P$ such that $f = h^{-1} \circ g \circ h$. Thus we have $f \circ f = (h^{-1} \circ g \circ h) \circ (h^{-1} \circ g \circ h)$. Thus $f \circ f = h^{-1} \circ g \circ (h \circ h^{-1}) \circ g \circ h = h^{-1} \circ g \circ i_A \circ g \circ h = h^{-1} \circ g \circ g \circ h$. Thus $f \circ fRg \circ g$.

**(c)**

Suppose $f(a) = a$ and $fRg$. Thus there exist an $h \in \mathcal P$ such that $f = h^{-1} \circ g \circ h$. Thus $f(a) = a = h^{-1} \circ g \circ h(a)$. Applying $h$ on both sides, $h(a) = g \circ h(a)$. Thus $g(h(a)) = h(a)$. Thus $g$ also has a fixed point = $h(a)$.