Let $f,g: [0,1] \to X$ be a continuous function ($[0,1]$ has the subspace topology), where $X$ is an arbitrary topological space.
Define $h: [0,1] \to X: t \mapsto \begin{cases}f(2t) \quad0 \leq t \leq 1/2 \\g(2t-1) \quad 1/2 \leq t \leq 1\end{cases}$
How can I quickly verify that $h$ is continuous?
EDIT: It must be the case that $f(1) = g(0)$, by definition of $h$
I cannot use limits, as these are metric space concepts and $X$ is not necessarily metrizable.
$t\rightarrow 2t$ is continuous for $0\leq t\leq 1/2$, and $f$ is continuous on $[0,1]$, then the composition $\varphi_{1}$ is continuous on $[0,1/2]$.
Similar reasoning is valid for $\varphi_{2}(t)=g(2t-1)$ on $[1/2,1]$.
$f$ is well-defined at $t=1/2$, and $[0,1/2],[1/2,1]$ are closed, by Gluing Lemma, $h$ is continuous:
\begin{align*} h^{-1}(F)&=\{t\in[0,1/2]: f(2t)\in F\}\cup\{t\in[1/2,1]: g(2t-1)\in F\}\\ &=\varphi_{1}^{-1}(F)\cup\varphi_{2}^{-1}(F) \end{align*} for closed sets $F$. Note that $\varphi_{1}^{-1}(F),\varphi_{2}^{-1}(F)$ are closed in $[0,1/2]$ and $[1/2,1]$ respectively, and hence closed in $[0,1]$.