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.
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Show that any continuous map $f:\partial D \to \partial D'$ can be extended to a continuous map $F:D \to D'$ such that $F(Int \; D) \subseteq Int \; D'$.
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Given points $p \in Int \; D$ and $p' \in Int \; D'$, show that $F$ can be chosen so $F(p) = p'$.
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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.
Best Answer
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. $\Box$