Closure of Bounded Component $\mathbb{E}^2−J$ in Armstrong's Basic Topology I'm trying to solve problem 42 from chapter 5 of M. A. Armstrong's Basic Topology:

Let $J$ be a polygonal Jordan curve in the plane, and let $X$ denote the closure
of the bounded component of $J$ ($\mathbb{E} ^2 - J$ ?). Show that $X$ can be broken up into a number
of convex regions by extending the edges of $J$, then divide each of these regions
into triangles. Now use induction on the number of triangles to show that $X$
is homeomorphic to a disc.

I can not figure out how tackle this problem since I have some intuitive ideas but I think they are special cases. On the other hand breaking up a closed set in plane to convex regions seems irrelevant to the topic. Also I'm confused about using induction for to show homeomorphism, they seem so irrelevant.
Any help is appreciated.
 A: The suggested solution has two steps: showing the division into triangles exists; and applying induction on the number of triangles.
Given your perception of irrelevance, I think that I will limit my answer to showing you how to do the second step, and I'll state some "clear" facts and lemmas about polyhedra and quotient topologies without proof. But let me know if you want further details about any of those issues.
Let me bridge the two steps with some definitions.

Define a polygonal Jordan region $X \subset \mathbb E^2$ to be the closure of the bounded component of $\mathbb E^2 - J$ for some polygonal Jordan curve $J$.


Define a triangulation of a polygonal Jordan region $X$ to be a finite collection of triangles $\{T_k\}_{k = 1}^K$ such that $A = \bigcup_{k=1}^K T_k$, and for each $k \ne k'$ the intersection $T_k \cap T_{k'}$ is either empty, or a common vertex of $T_k$ and $T_{k'}$, or a common edge of $T_k$ and $T_{k'}$.

So your two steps can be stated as follows:

Step 1: Every Jordan region $X$ has a triangulation.


Step 2: Given a triangulated Jordan region $X$ with $K$ triangles, show by induction on $K$ that $X$ is homeomorphic to a disc.

The basis step of the induction is when $K=1$ and so $X$ consists of a single triangle, and in this case $X$ is clearly homeomorphic to a disc.
The induction step is to assume that $K \ge 2$, and that every triangulated Jordan region with $<K$ triangles is homeomorphic to a disc, and use that to prove that $X$ is homeomorphic to a disc.
First an observation. Given an edge $E$ of the triangulation of $X$, either $E \subset J$ in which case there is a unique triangle containing $E$, or $E \not\subset J$ in which case there are exactly two triangles containing $E$.
Now the proof breaks into a few cases.
Case 1: There exists an edge $E$ of the triangulation of $X$ such that $E \not\subset J$ and both endpoints of $E$ are in $J$.
In this case $E$ subdivides $X$ into two Jordan regions $X_1$ and $X_2$, meaning that $X = X_1 \cup X_2$ and $X_1 \cap X_2 = E$.
The triangle on one side of $E$ belongs to $X_1$ and the triangle on the other side of $E$ belongs to $X_2$, and therefore the numbers of triangles of both $X_1$ and $X_2$ are strictly smaller than $K$. It follows, by the induction hypothesis, that $X_1$ and $X_2$ are each homemorphic to the disc.
We can now conclude that $X$ itself is homeomorphic to the disc by applying the following:

Lemma: If $X_1$, $X_2$ are two Jordan regions homeomorphic to the disc, and if $X_1 \cap X_2$ is a polygonal Jordan arc $A$, then $X_1 \cup X_2$ is homeomorphic to the disc.

This lemma is really a quotient topology exercise: one proves that $X_1 \cup X_2$ is the quotient of the disjoint union of $X_1$ and $X_2$ by identifying the copy of $A$ on the boundary of $X_1$ with the copy of $A$ on the boundary of $X_2$; and one uses this to prove that the quotient is homeomorphic to a disc.
Case 2: Every edge $E$ of the triangulation of $X$ such that $E \not\subset J$ has at most one endpoint on $J$. Choose any edge $E$ of the triangulation such that $E \subset J$. Let $T$ be the unique triangle of the triangulation which has $E$ as one of its edges. List the other two edges of $T$ as $E_1$ or $E_2$.
It is not possible that both of $E_1$ and $E_2$ is in $J$, and because in that case we would have $X=T$ and $K=1$, but we have assumed $K \ge 2$.
For whichever of $E_1$ or $E_2$ is not in $J$, it has one endpoint in $J$ and therefore its opposite endpoint is not in $J$, by the hypothesis of Case 2.
There are now two subcases:
Case 2a: One of $E_1$ or $E_2$ is in $J$, let's say $E_1 \subset J$. Then we have a subdivision of $X$ into two Jordan regions $X = X_1 \cup T$, $X_1 \cap T = E_2$. The triangulation of $X_1$ has one fewer triangle than $X$, so the induction hypothesis shows that $X_1$ is homeomorphic to a disc, and clearly $T$ is homeomorphic to a disc. The lemma implies that $X$ is homeomorphic to a disc.
Case 2b: Neither $E_1$ nor $E_2$ is in $J$. Since the common endpoint of $E_1$ and $E_2$ is not in $J$, it follows that $A = E_1 \cup E_2$ is a Jordan arc that separates $X$ into two Jordan regions $X = X_1 \cup X_2$, $X_1 \cap X_2 = A$. Again $X_1,X_2$ have fewer triangles, so induction and the lemma apply to prove that $X$ is homeomorphic to a disc.
