Extending a continuous map between the boundary of two cells.
I'm working in Lee's book on topological manifolds and have gotten stumped on the first question in chapter 5, the chapter on cell complexes. The problem is:
Let $D$ and $D'$ be two closed cells not necessarily of the same dimension.
Show that any continuous map $f:\partial D \to \partial D'$ can be extended to a continuous map $F:D \to D'$ such that $F(Int \; D) \subseteq Int \; D'$.
Given points $p \in Int \; D$ and $p' \in Int \; D'$, show that $F$ can be chosen so $F(p) = p'$.
Show that if $f$ is a homeomorphism then $F$ can be chosen to be a homeomorphism.
I proved the first part roughly as follows: First suppose $D$ and $D'$ are convex and each contain $0$ in their interior (in their respective ambient spaces). Then every element other than $0$ in $D$ can be expressed uniquely in the form $\lambda q$ where $q \in \partial D$ and $\lambda\in (0,1]$ (an equivalence relation on the chords connecting $0$ to boundary points partitions $D- \{0\}$). I then define the map $F(\lambda q) = \lambda f(q)$ which is continuous since if $\lambda_n q_n \to \lambda q$ in $D$ then $F(\lambda_n q_n) \to F(\lambda q)$. Lastly $\lambda q \in Int \; D$ implies $\lambda <1$ and so $\lambda f(q)$ is an interior point of $D'$ since $f(q)$ is a boundary point and $D'$ is convex.
Now if we suppose $D$ and $D'$ are arbitrary closed cells there are homeomorphisms $g_1: \overline{\mathbb{B}^n} \to D$ and $g_2: \overline{\mathbb{B}^m} \to D'$ (where possibly $m=n$) then we have that $g_2^{-1}\circ f \circ g_1$ is a continuous map between the boundaries of two closed balls and so by the first part can be extended to a continuous map $F:\overline{\mathbb{B}^n} \to \overline{\mathbb{B}^m}$. The mapping $g_2 \circ F \circ g_1^{-1}$ is a continuous map that satisfies the desired claim.
My issue is that parts 2 and 3 don't seem (to me at least) to follow that easily from the proof I constructed. Maybe my proof is wrong and I'm not seeing it, but more likely I think that I'm missing the spirit in which Lee is intending for us to approach this problem.
After some thought I was able to come to a solution for this problem. Thanks to those who commented above and gave me some hints, it was really helpful.
Part 2
By a proposition 5.1 in Lee's book, for any compact and convex $n$-cell $D$, and $p \in Int \; D$ there is a homeomorphism $g_p:\overline{\mathbb{B}^n} \to D$ where $g_p(0) = p, \; g_p(\mathbb{B}^n) = Int \; D$ and $g_p(S^{n-1}) = \partial D$. Starting with two arbitrary closed cells $D$ and $D',$ a continuous $f: \partial D \to \partial D'$, and $p \in Int \; D$ and $q \in Int \; D'$ there are homeomorphisms $g_p: \overline{\mathbb{B}^n} \to D$ and $g_q: \overline{\mathbb{B}^m} \to D'$ with the above property. $g_q^{-1}\circ f \circ g_p$ is a continuous map from the boundaries of two closed balls so from part 1 of the proof can be extended continuously to $F:\overline{\mathbb{B}^n} \to \overline{\mathbb{B}^m}$. The map $g_q \circ F \circ g_p^{-1}$ is continuous, preserves the map $f$ and
$$ (g_q\circ F \circ g_p^{-1})(p) \;\; =\;\; g_q(F(0)) \;\; =\;\; g_q(0) \;\; =\;\; q $$
where $F$, as constructed from part 1, satisfies $F(0) = 0$. $\Box$
Part 3
Suppose $D$ and $D'$ are convex with $0$ in their respective interiors, $f:\partial D\to \partial D'$ a homeomorphism, and use the same map $F:D \to D'$ namely $F(\lambda q) = \lambda f(q)$. $F$ is injective since if $\lambda f(q) = \gamma f(p)$ and either $\gamma$ or $\lambda =0$, then the other scalar must also be zero since $f(q), f(p) \neq 0$. If neither scalar is zero then $f(p) = \frac{\lambda}{\gamma}f(q)$ which means $f(p)$ and $f(q)$ lie on the same chord in $D'$. This only occurs when $f(p) = f(q)$ since they are both boundary points, and since $f$ is bijective $p=q$. This implies $f(p) = \frac{\lambda}{\gamma} f(p)$ hence $\lambda = \gamma$ so $F$ is injective. Surjectivity follows since any point in $D'$ can be written $\lambda m$ for $m \in \partial D'$, which has a unique image point $f^{-1}(m) \in \partial D$. We then have $F(\lambda f^{-1}(m)) = \lambda m$ hence $F$ is bijective. Since $F$ is continuous, bijective, and maps a compact space into a Hausdorff space then by the closed map lemma it is a homeomorphism.
For arbitrary closed cells $D$ and $D'$ the proof is similar to those given above using homeomorphisms $g_1: \overline{\mathbb{B}^n} \to D$ and $g_2:\overline{\mathbb{B}^m} \to D'$. $g_2^{-1}\circ f\circ g_1$ is a homeomorphism between the boundaries of two closed balls and so is extended to a homeomorphism $F$ between the balls. Then $g_2\circ F \circ g_1^{-1}$ is the desired homeomorphism. $\Box$