How to show that the composition of two paths is contiuous Let ($X,\tau$) be a topological space. Let $x,y,z \in X$.
Let $\alpha \in P_{(X,\tau)}(x,y)$ and $\beta \in P_{(X,\tau)}(y,z)$. How do I show that the meld $\alpha * \beta$ of $\alpha$ and $\beta$, defined as:
$\alpha * \beta (s) = \begin{cases}
       \alpha(2s) &s \in [0,\frac{1}{2}]\\
       \beta(2s-1) &s \in [\frac{1}{2},1]\\
     \end{cases} $
is continuous? I know about the glue lemma but in order to apply this I have to show that $\alpha * \beta _{|[0,\frac{1}{2}]}$ and $\alpha * \beta _{|[\frac{1}{2},1]}$ are continuous. Even though I know $\alpha$ and $\beta$ are continuous and it seems "obvious" that this would make $\alpha * \beta _{|[0,\frac{1}{2}]}$ and $\alpha * \beta _{|[\frac{1}{2},1]}$ also continuous, but what is the exact  argument to show this?
 A: To close the question: define $h_1: [0,\frac12] \to [0,1]$ by $h_1(s)=2s$. It is clear that $h_1$ is continuous (even a homeomorphism).
Also define $h_2: [\frac12, 1] \to [0,1]$ by $h_2(s)=2s-1$. Also continuous and a homeomorphism. (both can be shown metrically by noting that $\delta=\frac{\varepsilon}{2}$ works uniformly, if you must; they also are increasing bijections so that's also a possible argument for the order topology).
Then note that $(\alpha \ast \beta)\restriction_{[0,\frac12]} = \alpha \circ h_1$ which is continuous as a composition of two continuous functions.
Same for $(\alpha \ast \beta)\restriction_{[\frac12,1]} = \beta \circ h_2$.
And as both $(\alpha \ast \beta)\restriction_{[0,\frac12]}(\frac12)= \alpha(1)=y$ and $(\alpha \ast \beta)\restriction_{[\frac12,1]}(\frac12)=\beta(0)=y$, they agree on the overlap $[0,\frac12] \cap [\frac12,1] = \{\frac12\}$ so the glueing lemma applies as both parts are closed in $[0,1]$.
That should be detailed enough. Of course in practice such things are never written out in such boring detail..
