Let $X$ be a topological space. If $\gamma_1,\gamma_2:[0,1]\rightarrow X$ are continuous functions and $\gamma_1(1)=\gamma_2(0)$, show that $$\gamma:[0,1]\rightarrow X, \gamma(t)= \begin{cases} \gamma_1(2t) &\mbox{ if } t \in[0,\frac{1}{2}) \\ \gamma_2(2t-1) &\mbox{ if } t\in(\frac{1}{2},1] \end{cases}$$

is continuous.

I think this has something to do with the Urysohn lemma, because $[0,1]$ is compact and Hausdorff, but I'm not sure how this could help me.

  • $\begingroup$ No, it has nothing to do with Urysohn. It's trivial from the definition of "continuous". $\endgroup$ – David C. Ullrich Jan 23 '16 at 13:54

It's the pasting lemma you need: We have two continuous functions $f_1$ and $f_2$ defined on closed subsets $C_1$ and $C_2$ of a space $X$, such that $f_1(x) = f_2(x)$ for all $x \in C_1 \cap C_2$. Then $f$, the combination function, is continuous on $C = C_1 \cup C_2$. Here $C_1 = [0,\frac{1}{2}]$ and $C_2 = [\frac{1}{2},1]$. No need for compactness etc.

See Wikipedia and this question, e.g.

  • $\begingroup$ Thank you! I've got one question left. In this case, $C_2=(\frac{1}{2},1]$, an half open interval. I know this is locally compact, but not closed.. So does the pasting lemma still 'work' then? $\endgroup$ – jbuser430 Jan 23 '16 at 14:00
  • $\begingroup$ No $C_2$ is $[\frac{1}{2},1]$. Which is closed. On the intersection, $\frac{1}{2}$, you should check that the two functions coincide. Then the lemma applies. $\endgroup$ – Henno Brandsma Jan 23 '16 at 14:01
  • $\begingroup$ ah, I see. Thanks again! $\endgroup$ – jbuser430 Jan 23 '16 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.