Proving that the empty glass is a deformation retract of the full glass

Question

Let me write down the exercise described in the title formally.

Let $X=E^2 \times I$ and $A=E^2 \times {0} \cup S^1 \times I$, where $E^2$ designates the ball of radius 1 in $\mathbb{R}^2$.

I want to show that $X$ is a deformation retract of $A$.

I must give a continous function $F:X\times I\rightarrow X$ that meets the 3 requirements below:

1. $F(x,0)\in A$ for all $x\in X$;
2. $F(x,1)=x$ for all $x\in X$;
3. $F(a,t)=a$ for all $a\in A$ and for all $t\in I$.

Let $F(x,t)=t\cdot x+(1-t)\cdot\theta(x)$, where $\theta(x)$ is the intersection between $A$ and the line that unites $x$ to the point $(0,0,2)$. It's easy to see that $F$ meets the conditions above.

My question is how do I reason that this function is indeed continuous? Hatcher (Proposition 0.16) mentions something about the continuity of this map in a language completely foreign to me (due to my null knowledge of algebraic topology).

Answer 1

While @String gave you the explicit formula for $\theta$ I will show you how to generalize it to any "good enough" closed subset and a projection onto it.

So what is $\theta$?

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More generally consider a Banach space $E$, a fixed vector $w\in E$ and a closed subset $A\subseteq E$ such that $w\not\in A$. Now for any $v\in E$ consider following sets

$$I_v=\big\{tv+(1-t)w\ \big|\ t\in\mathbb{R}\big\}$$ $$A_v=I_v\cap A$$ $$B=\{v\in E\ |\ A_v\neq\emptyset\}$$

and the multivalued function

$$f:B\multimap E$$ $$f(v)=A_v$$

I leave as an exercise that $f$ is lower semicontinuous.

Now you can easily check that $A_v$ is closed. Assume that $A_v$ is additionally convex for each $v\in B$. Since $B$ is paracompact (as a metric space) the Michael selection theorem applies to $f$ and thus there exists a continuous selection $s:B\to E$ (i.e. $s(v)\in f(v)$).

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In our case $E=\mathbb{R}^3$, $w=(0,0,2)$ and $A=E^2\times\{0\}\cup S^1\times I$. You can easily check that $A_v$ has exactly one point if non-empty. Thus $A_v$ is convex and by the Michael selection theorem there is a continuous selection. Note that this selection is unique (simply because $A_v$ has one point). The restriction of that selection to $X$ is equal to $\theta$.

Answer 2

To prove that $\theta$ is continuous, the idea is to convert the geometric description given into a formula for $\theta$, and then to apply that formula. You'll be using what you know from calculus and analytic geometry, with a bit of topology, to prove that $\theta$ is continuous (so it's not algebraic topology that you use to prove continuity).

But, there is a twist: the formula will be in two pieces, corresponding to the two pieces in the expression $$A = (E^2 \times 0) \cup (S^1 \times I) $$ So, the outline for the proof goes like this:

1. Write down set theoretic formulas for decomposing the domain $X$ into two parts $X = X_1 \cup X_2$, such that $$X_1 = \theta^{-1}(E^2 \times 0) $$ and $$X_2 = \theta^{-1}(S^1 \times I) $$
2. Write down formulas for $\theta | X_1$ and $\theta | X_2$. These should be ordinary formulas which you can verify continuity of, using ordinary tools of calculus.
3. To prove continuity of $\theta$ itself, apply the pasting lemma: verify that $X_1 \cap X_2$ is closed, and that the formulas for $\theta | X_1$ and $\theta | X_2$ agree on the intersection $X_1 \cap X_2$, and you are done.

Carrying out steps 1 and 2 requires good analytic geometry skills. I'll get you started by carrying out steps 1 and 2 for $X_1$, leaving you to try $X_2$ (which is harder).

The set $X_1$ is a frustum, obtained by starting with the solid circular cone having base $E^2 \times 0$ and apex $(0,0,2)$, and then intersecting that cone with the solid cylinder $E^2 \times I$. You can write it as $$X_1 = \{(x,y,z)\quad \,|\,\quad x^2 + y^2 \le \left(1-\frac{z}{2}\right)^2 \quad\text{and}\quad 0 \le z \le 1 \} $$ To get the formula for $\theta(x,y,z)$ when $(x,y,z) \in X_1$, take the ray from $(0,0,2)$ through $(x,y,z)$, and intersect that ray with $E^2 \times \{0\}$. You get $$\theta(x,y,z) = (0,0,2) + \frac{2}{2-z}\bigl((x,y,z)-(0,0,2)\bigr) = \bigl(\frac{2x}{2-z},\frac{2y}{2-z},0\bigr) $$ Since the plane $z=2$ is disjoint from the set $X_1$, this formula shows that $\theta | X_1$ is continuous. (You might also want to verify that this formula has image in $E^2 \times 0$ when $(x,y,z) \in X_1$).

Here, very briefly, are some portions of steps 1 and 2 of $X_2$. The set $X_2$ is the closure of the complement of frustrum $X_1$ inside the cylinder $E^2 \times I$, and so $$X_2 = \{(x,y,z) \quad \,|\, \quad (1-\frac{z}{2})^2 \le x^2 + y^2 \le 1 \quad\text{and}\quad 0 \le z \le 1\} $$ The formula for $\theta(x,y,z)$ will instead be obtained by taking the ray from $(0,0,2)$ through $(x,y,z)$ and intersecting with the cylindrical surface $S^1 \times I$.

Finally, you should be able to write down the set theoretic formula for $X_1 \cap X_2$ and verify that it is a closed set, and that the two parts of the formula for $\theta$ give the same outcome for $(x,y,z) \in X_1 \cap X_2$.

Answer 3

Regarding $\theta(\mathbf x)$ where $\mathbf x=(x,y,z)$ we have $$ \theta(\mathbf x)= \begin{cases} \frac{1}{\sqrt{x^2+y^2}}(x,y,z-2)+ (0,0,2) &\text{if}\quad \sqrt{x^2+y^2}>\frac{2-z}{2}\\ \quad\\ \frac{2}{2-z}(x,y,0) &\text{otherwise} \end{cases} $$

if I am not mistaken.

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These expressions are continuous since when $$ \sqrt{x^2+y^2}=\frac{2-z}{2} $$ we have $$ \begin{align} \frac{1}{\sqrt{x^2+y^2}}(x,y,z-2)+ (0,0,2) &= \frac{2}{2-z}(x,y,z-2)+ (0,0,2)\\ &= \frac{2}{2-z}(x,y,0) \end{align} $$ and all the rest is just compositions, sums, and products of continuous functions.