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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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