Problem:
Let $f$ be a path from $a$ to $b$. Show that $g$ defined by
$g(x)= \left\{ \begin{array}{ll} f(2x) & x \in [0, \frac{1}{2}] \\ b & x\in[\frac{1}{2}, 1] \\ \end{array} \right. $
is path-homotopic to $f$.
My Solution:
Define $H : I \times I \rightarrow X$,
$H(x, t)= \left\{ \begin{array}{ll} f(x+tx) & x \in [0, \frac{1}{1+t}] \\ b & x\in[\frac{1}{1+t}, 1] \\ \end{array} \right. $
We can see that at $x = \frac{1}{1+t}$,
$f(\frac{1}{1+t}+\frac{t}{1+t}) = f(1) = b$ (as $f$ is a path from $a$ to $b$)
Also, (checking criteria for homotopy),
$H(x,0)= f(x)$
$H(x, 1)= \left\{ \begin{array}{ll} f(2x) & x \in [0, \frac{1}{2}] \\ b & x\in[\frac{1}{2}, 1] \\ \end{array} \right. = g(x)$
$H(0,t)= f(0) = a$
$H(1,t)= g(1)=b$
So, I do believe I have defined a homotopy. However, I have been asked to show continuity. I thought that because $f$ is continuous and $b$, a constant, is continuous, then $H$ would be continuous. Though, my professor said that is not enough to prove continuity... that it only shows continuity along one line. So my question is...
- How can continuity fail?
- How do I prove continuity in another way? (Perhaps $\epsilon - \delta$?)
You can always glue two continous functions defined on closed subsets to a continous function provided that they agree on the intersection. This is the the Pasting lemma.
Here the subsets are $A=\{(x,t)\in I\times I : x\leq\frac{1}{1+t}\}$ and $B=\{(x,t)\in I\times I : x\geq \frac{1}{1+t}\}$.
$A$ and $B$ are closed since they are the preimages of $\mathbb R_{\leq 0}$ and $\mathbb R_{\geq 0}$ under the continous map $(x,t)\mapsto x-\frac{1}{1+t}$. $H_{|A}$ is continous as a composition of continous maps
$$H_{|A}: A\xrightarrow{(x,t)\mapsto x+tx}I\xrightarrow{f}X$$
and $H_{|B}\equiv b$ is continous as a constant map.
Now i think was your professor meant was that you only showed that for fixed $t$ the function $x\mapsto H(x,t)$ is continous. In general this does not imply that $H$ is continous, consider for example $H(x,t)=d(t)$ where $d:I\to X$ is any discontinous function.