diff --git a/Bookshelf/Enderton/Set.tex b/Bookshelf/Enderton/Set.tex
index 2cfd8a8..f1eee7d 100644
--- a/Bookshelf/Enderton/Set.tex
+++ b/Bookshelf/Enderton/Set.tex
@@ -44,7 +44,7 @@ For any set $I$ and any function $H$ with domain $I$, if $H(i) \neq \emptyset$
 
 \end{axiom}
 
-\section{\pending{Cartesian Product}}%
+\section{\defined{Cartesian Product}}%
 \label{ref:cartesian-product}
 
 Let $I$ be a set and let $H$ be a \nameref{ref:function} whose domain includes $I$.
@@ -54,6 +54,12 @@ We define the \textbf{cartesian product} of the $H(i)$'s as
     f \text{ is a function with domain } I \text{ and }
       (\forall i \in I) f(i) \in H(i)\}.$$
 
+\begin{definition}
+
+  \lean*{Mathlib/Data/Set/Prod}{Set.prod}
+
+\end{definition}
+
 \section{\defined{Compatible}}%
 \label{ref:compatible}
 
@@ -2974,12 +2980,11 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
       All that remains is to prove $F$ is single-rooted.
       Let $y \in \ran{F}$.
       By definition of the \nameref{ref:range} of a function, there exists some
-        $x$ such that $\left< x, y \right> \in F$.
-      Suppose $x_1, x_2 \in \dom{F}$ such that
-        $\left< x_1, y \right>, \left< x_2, y \right> \in F$.
-      Then $F(x_1) = F(x_2) = y$.
-      Then $G(F(x_1)) = G(F(x_2))$ implies $I_A(x_1) = I_A(x_2)$.
+        $x_1$ such that $\left< x_1, y \right> \in F$.
+      Suppose there exists a set $x_2$ such that $\left< x_2, y \right> \in F$.
+      By hypothesis, $G(F(x_1)) = G(F(x_2))$ implies $I_A(x_1) = I_A(x_2)$.
       Thus $x_1 = x_2$.
+      Therefore $F$ must be single-rooted.
 
     \subparagraph{($\Leftarrow$)}%
 
@@ -4663,7 +4668,7 @@ Show that $G$ maps $\powerset{A}$ one-to-one into $\powerset{B}$.
 
 \end{proof}
 
-\subsection{\pending{Exercise 3.29}}%
+\subsection{\verified{Exercise 3.29}}%
 \label{sub:exercise-3.29}
 
 Assume that $f \colon A \rightarrow B$ and define a function
@@ -4677,6 +4682,9 @@ Does the converse hold?
 
 \begin{proof}
 
+  \lean{Bookshelf/Enderton/Set/Relation}
+    {Enderton.Set.Chapter\_3.exercise\_3\_29}
+
   Let $f \colon A \rightarrow B$ such that $f$ maps $A$ onto $B$.
   Define $G \colon B \rightarrow \powerset{A}$ by
     \eqref{sub:exercise-3.29-eq1}.
@@ -4693,7 +4701,7 @@ Does the converse hold?
       G(x_2) & = \{x \in A \mid f(x) = x_2\}.
     \end{align*}
   Since $f$ maps $A$ onto $B$, $\ran{f} = B$.
-  Thus $x_2, x_2 \in \ran{f}$.
+  Thus $x_1, x_2 \in \ran{f}$.
   By definition of the \nameref{ref:range} of a set, there exist some $t \in A$
     such that $f(t) = x_1$.
   Therefore $t \in G(x_1)$.
diff --git a/Bookshelf/Enderton/Set/Chapter_3.lean b/Bookshelf/Enderton/Set/Chapter_3.lean
index bfdc161..9d7d183 100644
--- a/Bookshelf/Enderton/Set/Chapter_3.lean
+++ b/Bookshelf/Enderton/Set/Chapter_3.lean
@@ -508,19 +508,19 @@ Assume that `F : A → B`, and that `A` is nonempty. There exists a function
 **iff** `F` is one-to-one.
 -/
 theorem theorem_3j_a {F : Set.HRelation α β} {A : Set α} {B : Set β}
-  (hF : isSingleValued F ∧ mapsInto F A B) (hA : Set.Nonempty A)
+  (hF : mapsInto F A B) (hA : Set.Nonempty A)
   : (∃ G : Set.HRelation β α,
       isSingleValued G ∧ mapsInto G B A ∧
-        (∀ p ∈ comp G F, p.1 = p.2)) ↔ isOneToOne F := by
+        (comp G F = { p | p.1 ∈ A ∧ p.1 = p.2 })) ↔ isOneToOne F := by
   apply Iff.intro
-  · intro ⟨G, ⟨hG₁, hG₂, hI⟩⟩
+  · intro ⟨G, hG⟩
     refine ⟨hF.left, ?_⟩
-    show isSingleRooted F
     intro y hy
-    have ⟨x, hx⟩ := ran_exists hy
-    sorry
-  · intro h
+    have ⟨x₁, hx₁⟩ := ran_exists hy
+    refine ⟨x₁, ⟨mem_pair_imp_fst_mem_dom hx₁, hx₁⟩, ?_⟩
+    intro x₂ hx₂
     sorry
+  · sorry
 
 /-- #### Theorem 3J (b)
 
@@ -529,10 +529,10 @@ Assume that `F : A → B`, and that `A` is nonempty. There exists a function
 `B` **iff** `F` maps `A` onto `B`.
 -/
 theorem theorem_3j_b {F : Set.HRelation α β} {A : Set α} {B : Set β}
-  (hF : isSingleValued F ∧ mapsInto F A B) (hA : Set.Nonempty A)
+  (hF : mapsInto F A B) (hA : Set.Nonempty A)
   : (∃ H : Set.HRelation β α,
       isSingleValued H ∧ mapsInto H B A ∧
-        (∀ p ∈ comp F H, p.1 = p.2)) ↔ mapsOnto F A B := by
+        (comp F H = { p | p.1 ∈ B ∧ p.1 = p.2 })) ↔ mapsOnto F A B := by
   sorry
 
 /-- #### Theorem 3K (a)
@@ -1642,6 +1642,51 @@ theorem exercise_3_28 {A : Set α} {B : Set β}
     have hz := mem_pair_imp_snd_mem_ran hb.right
     exact hf.right.right.right hz
 
+/-- #### Exercise 3.29
+
+Assume that `f : A → B` and define a function `G : B → 𝒫 A` by
+```
+G(b) = {x ∈ A | f(x) = b}
+```
+Show that if `f` maps `A` *onto* `B`, then `G` is one-to-one. Does the converse
+hold?
+-/
+theorem exercise_3_29 {f : Set.HRelation α β} {G : Set.HRelation β (Set α)}
+  {A : Set α} {B : Set β} (hf : mapsOnto f A B)
+  (hG : mapsInto G B (𝒫 A) ∧ G = { p | p.1 ∈ B ∧ p.2 = {x ∈ A | (x, p.1) ∈ f} })
+  : isOneToOne G := by
+  unfold isOneToOne
+  refine ⟨hG.left.left, ?_⟩
+  intro y hy
+  have ⟨x₁, hx₁⟩ := ran_exists hy
+  refine ⟨x₁, ⟨mem_pair_imp_fst_mem_dom hx₁, hx₁⟩, ?_⟩
+  intro x₂ hx₂
+
+  have hG₁ : (x₁, {x ∈ A | (x, x₁) ∈ f}) ∈ G := by
+    rw [hG.right, ← hG.left.right.left]
+    simp only [Set.mem_setOf_eq, and_true]
+    exact mem_pair_imp_fst_mem_dom hx₁
+  have hG₂ : (x₂, {x ∈ A | (x, x₂) ∈ f}) ∈ G := by
+    rw [hG.right, ← hG.left.right.left]
+    simp only [Set.mem_setOf_eq, and_true]
+    exact hx₂.left
+  have heq : {x ∈ A | (x, x₁) ∈ f} = {x ∈ A | (x, x₂) ∈ f} := by
+    have h₁ := single_valued_eq_unique hG.left.left hx₁ hG₁
+    have h₂ := single_valued_eq_unique hG.left.left hx₂.right hG₂
+    rw [← h₁, ← h₂]
+
+  rw [hG.right, ← hf.right.right] at hG₁ hG₂
+  simp only [Set.mem_setOf_eq, and_true] at hG₁ hG₂
+  have ⟨t, ht⟩ := ran_exists hG₁
+  have : t ∈ {x ∈ A | (x, x₁) ∈ f} := by
+    simp only [Set.mem_setOf_eq]
+    refine ⟨?_, ht⟩
+    rw [← hf.right.left]
+    exact mem_pair_imp_fst_mem_dom ht
+  rw [heq] at this
+  simp only [Set.mem_setOf_eq] at this
+  exact single_valued_eq_unique hf.left this.right ht
+
 end Relation
 
 end Enderton.Set.Chapter_3
\ No newline at end of file