#import "@youwen/zen:0.1.0": *
#import "@preview/cetz:0.3.2"

#show: zen.with(
  title: "Homework 8",
  author: "Youwen Wu",
)

#set heading(numbering: none)
#show heading.where(level: 2): it => [#it.body.]
#show heading.where(level: 3): it => [#it.body.]
#set par(first-line-indent: 0pt)
#show figure: it => {
  pad(
    y: 10pt,
    it,
  )
}

Problems:

- 4.2: \#1g, 2b, 3b, 6, 7, 8, 14bd

- 4.3: \#1bdeh, 2bdeh, 3ab, 5, 6, 7, 8

= 4.2

== 1

=== g

$f(x) = 1 / (1-x)$

$f^(-1) (x) = 1 - 1 / x$

== 2

=== b

$f(x) = x^2 + 2x$, $g(x) = 2x + 1$.

$
  f compose g &= (2x+1)^2 + 2(2x+1) = 4x^2 + 8x + 3 \
  g compose f &= 2(x^2 + 2x) + 1 = 2x^2 + 4x + 1
$

== 3

=== b

$"Dom"(f compose g) = RR$, $"Dom"(g compose f = RR$.

$"Rng"(f compose g) = [3,infinity]$, $"Rng"(g compose f) = [1, infinity]$.

== 6

#theorem[
  If $f : A -> B$, then $I_B compose f = f$.
]

#proof[
  $"Dom"(I_B compose f) = "Dom"(f) = A$. Now suppose $x in A$, then $(I_B
compose f)(x) = I_B (f(x)) = f(x)$. Therefore $I_B compose f = f$.
]

== 7

#theorem[
  Let $f : A -> B$ with $"Rng"(f) = C$. If $f^(-1)$ is a function, $f compose f^(-1) = I_C$.
]

#lemma[
  $"Dom"(f^(-1)) = "Rng"(f)$.

  #proof[
    Suppose $(x,y) in f$. Then $(y,x) in f^(-1)$. Therefore for any $x$ in $"Dom"(f)$, $x$ is also in $"Rng"(f^(-1))$, so $"Dom"(f) subset.eq "Rng"(f^(-1))$.

    Suppose $(x,y) in f^(-1)$. Then $(y,x) in f$. So for any $y in "Rng"(f^(-1))$, $y$ is also in $"Dom"(f)$. So $"Rng"(f^(-1)) subset.eq "Dom"(f)$. Therefore $"Dom"(f) = "Rng"(f^(-1))$.
  ]
]<dom-inverse-rng>

#proof[
  Suppose that $f : A -> B$ and $f^(-1)$ is a function. By @dom-inverse-rng, $"Dom"(f^(-1)) = "Rng"(f) = "Dom"(I_C) = C$.

  Suppose that $y in C$. Because $(y,f^(-1) (y)) in f^(-1)$, $(f^(-1) (y),y) in f$. Then $f compose f^(-1) = f(f^(-1) (y)) = y = I_C$.
]

== 8

The graph of $lr(f|)_A$ is simply the collection of points ${(1,1),(2,-2),(3,-5),(4,-8)}$

The graph of $lr(f|)_[-1,3]$ is simply the line given by $f(x) = 4 - 3x$
restricted between the $x$-coordinates $[-1,3]$.

The graph of $lr(f|)_(2,4]$ is the line given by $f(x) = 4 - 3x$ between $x =
2$ to $x = 4$, with a hole at $x = 2$.

The graph of $lr(f|)_{6}$ is simply the point at $(6,-14)$.

== 14

=== b

$
  &h : [-1, infinity) -> RR, &&h(x) = x^2 + 1 \
  &g : (-infinity, -1] -> RR, &&g(x) = x + 3
$

$h union g$ is a function only if $h(x) = g(x)$ where their domain overlaps at
-1. We check $h(-1) = g(-1) = 2$ is indeed true. Everywhere else $h union g$ is
either $g(x)$ or $h(x)$ which are both functions. Therefore it is a function.

=== d

$
  &h : (-infinity,2] -> RR, &&h(x) = cos x \
  &g : [2, infinity) -> RR, &&g(x) = x^2
$

We check that $h$ and $g$ agree at their overlapping domain. $h(2) = cos 2$,
but $g(2) = 4$. Without further computation note that $cos(x)$ gives values
only in $[-1,1]$ so $h(2) != g(2)$. $h union g$ would contain both $(2,4)$ and
$(2,cos 2)$ so it's not a function.

= 4.3

== 1

=== b

Suppose $x in ZZ$. Let $w = -x + 1000$. We note that $w in "Dom"(f)$. Then
$f(w) = -(-x + 1000) + 1000 = x$. Hence $f$ is onto $RR$.

=== d

Let $x in RR$. Let $w = root(3,x)$. Then $f(w) = (root(3,x))^3 = x$.
Additionally, we note $w in RR$. Hence $f$ is onto $RR$.

=== e

Attempt to find $x$ such that $f(x) = -2$. Note that $f$ only produces positive
real values due to the square root. Hence no such $x$ can exist. However $-2 in
RR$ so $f$ is not onto $RR$.

=== h

Suppose $z in RR$. Let $w = (z,0)$. Note that $w in R times R$. Then $f(w) =
f(z,0) = z$. Hence $f$ is onto $RR$.

== 2

=== b

Suppose that $f(x) = f(z)$ for $x,z in ZZ$. Then $-x + 1000 = -z + 1000$. After
manipulation, we see that $x = z$. Hence $f$ is one-to-one.

=== d

Suppose that $f(x) = f(z)$ for $x,z in RR$. Then $x^3 = z^3$. Then $root(3,x^3)
= root(3,z^3)$ and $x = z$. So $f$ is one-to-one.

=== e

Consider $f(2) = sqrt(9) = 3$ and $f(-2) = sqrt(9) = 3$. 3 has both $2$ and
$-2$ in its preimage. So it's not one-to-one.

=== h

Consider $f(2,1) = 2 - 1 = 1$ and $f(3,2) = 3 - 2 = 1$. Therefore 1 has both
$(2,1)$ and $(3,2)$ in its preimage and it is not one-to-one.

== 3

=== a

Suppose $x in A$. Then $I_A (x) = x$. Since we said $x in A$, $x$ always has $x$ in its preimage. So $I_A$ is onto $A$.

Now suppose that $x,z in A$ and $I_A (x) = I_A (z)$. Then $x = z$ so $I_A$ is
one-to-one.

=== b

Suppose $overline(x) in ZZ_5$ (here $overline(x)$ is a representative of the
equivalence class containing $x$). Note that $x in ZZ$. Then $f(x) =
overline(x)$, so $f$ is onto $ZZ_5$. Now consider that $f(3)$ and $f(8)$. Then
$3 = 8 space (mod 5)$, hence $f(3) = f(8) = overline(3)$. So $f$ is not
one-to-one.

== 5

#theorem[
  If $f : A -> B$ and $g : B -> C$ is onto $C$, then $g compose f$ is onto $C$.
]

#proof[
  Suppose $x in C$. Then there exists $y in B$ such that $g(y) = x$.
  Additionally, there exists $z in A$ such that $f(z) = y$. Then $(g compose
  f)(z) = g(f(z)) = g(y) = x$. So if $x in C$, there is always $z in A$ such
  that $(g compose f)(z) = x$. Therefore $g compose f$ is onto $C$.
]

== 6

#theorem[
  Let $f : A -> B$ and $g : B -> C$. If $g compose f$ is one-to-one, then $f$
  is one-to-one.
]

#proof[
  Assume $f(x) = f(z)$. Then $g(f(x)) = g(f(z))$ because $g$ is a function.
  Then $(g compose f)(x) = (g compose f)(z)$. Because $g compose f$ is
  one-to-one, it follows that $x = z$. Therefore $f$ is one-to-one.
]

== 7

#theorem[
  If $f : A -> B$ is one-to-one, then every restriction of $f$ is one-to-one.
]

#proof[
  Let $D subset.eq A$. Then consider the restriction $lr(f)|_D$. Suppose
  $lr(f|)_D (x) = lr(f|)_D (z)$, for $x$ and $z$ in $D$. Then $x$ and $z$ are
  in $A$, and $lr(f|)_D (x) = lr(f|)_D (z) = f(x) = f(z)$ so $x = z$ by the
  injectivity of $f$. Hence $lr(f|)_D$ is still one-to-one.
]

== 8

We're proving the following:

Let $h : A -> C$ and $g : B -> D$ be functions.

If $A$ and $B$ are disjoint, $h$ is onto $C$, and $g$ is onto $D$, then $h
union g : A union B -> C union D$ is onto $C union D$.

#proof[
  First note that $h union g$ is indeed a function by a result we proved
  earlier in the homework.

  Suppose $x in C union D$. Then $x in C$ or $x in D$ (here we are using the
  inclusive or). Then at least one of the following must be true, by the
  surjectivity of $h$ and $g$:

  + We can find $z in A$ such that $h(z) = x$.
  + We can find $z in B$ such that $g(z) = x$.

  In either case, if $z in A$ then $z in A union B$ and if $z in B$ then $z in
  A union B$. So we found some $z in A union B$ such that $(h union g)(z) = x$.
  Therefore $h union g$ is onto $C union D$.
]