auto-update(nvim): 2025-03-06 01:36:57

documents/by-course/math-8/course-notes/main.typ

@@ -1348,7 +1348,7 @@ function.
 
 #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(f(x)) = (g
+  Additionally, there exists $z in A$ such that $f(z) = y$. Then $g(f(z)) = (g
   compose f)(z) = x$. So if $x in C$, there is always $z in A$ such that $(g
   compose f)(z) = x$. Therefore $g compose f$ is onto $C$.
 ]

documents/by-course/math-8/pset-8/main.typ

@@ -23,8 +23,6 @@ Problems:
 
 - 4.3: \#1bdeh, 2bdeh, 3ab, 5, 6, 7, 8
 
-#outline()
-
 = 4.2
 
 == 1
@@ -123,3 +121,131 @@ 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 $C union D$.
+]