Prove that $[0,\frac{1}{2}]\ni x\mapsto 2x\in [0,1]$ is continuous.

Question

Let $S$ be a topological space.
Let $f$ be a continuous mapping from the topological subspace $[0,1]$ of $\mathbb{R}$ to $S$.
Let $f_1$ be the mapping from the topological subspace $[0,\frac{1}{2}]$ of $\mathbb{R}$ to $S$ such that $f_1(x):=f(2x)$.
Prove that $f_1$ is a continuous mapping from the topological subspace $[0,\frac{1}{2}]$ of $\mathbb{R}$ to $S$.

My solution is here:

Let $g$ be a mapping from the topological subspace $[0,\frac{1}{2}]$ of $\mathbb{R}$ to the topological subspace $[0,1]$ of $\mathbb{R}$ such that $g(x)=2x$.
Then, $f_1=f\circ g$.
Since $f$ is continuous, we need only to prove $g$ is continuous.
We need to prove if $O'$ is an open subset of $\mathbb{R}$, then $g^{-1}([0,1]\cap O')$ is open in the topological subspace $[0,\frac{1}{2}]$ of $\mathbb{R}$.
$g^{-1}([0,1]\cap O')$ is open in the topological subspace $[0,\frac{1}{2}]$ of $\mathbb{R}$ if and only if there is an open subset $O$ of $\mathbb{R}$ such that $g^{-1}([0,1]\cap O')=[0,\frac{1}{2}]\cap O$.

Let $g_1$ be a mapping from the topological space $\mathbb{R}$ to the topological space $\mathbb{R}$ such that $g_1(x):=2x$.
Then, $g_1$ is a continuous mapping.
Proof:
Let $O'$ be an open subset of $\mathbb{R}$.
Then, $g_1^{-1}(O')=\{x\in\mathbb{R}\mid 2x\in O'\}$.
Let $x_0\in g_1^{-1}(O')$.
Then, $2x_0\in O'$.
Since $O'$ is open in $\mathbb{R}$, there exists a positive real number $\epsilon$ such that $(2x_0-\epsilon,2x_0+\epsilon)\subset O'$.
Let $x\in (x_0-\frac{\epsilon}{2},x_0+\frac{\epsilon}{2})$.
Then, $x_0-\frac{\epsilon}{2}<x<x_0+\frac{\epsilon}{2}$ holds.
So, $2x_0-\epsilon<2x<2x_0+\epsilon$ holds.
So, $2x\in (2x_0-\epsilon,2x_0+\epsilon)\subset O'$.
So, $x\in g_1^{-1}(O')$.
Therefore, $(x_0-\frac{\epsilon}{2},x_0+\frac{\epsilon}{2})\subset g_1^{-1}(O')$.
So, $g_1^{-1}(O')$ is an open subset of $\mathbb{R}$.
So, $g_1$ is a continuous mapping.

Let $O'$ be an open subset of $\mathbb{R}$.
Then, $g^{-1}([0,1]\cap O')=[0,\frac{1}{2}]\cap g_1^{-1}(O')$ holds.
Proof:
Let $x_0\in g^{-1}([0,1]\cap O')$.
Then, $x_0\in[0,\frac{1}{2}]$ since the domain of $g$ is $[0,\frac{1}{2}]$.
Since $g(x)=g_1(x)$ on $[0,\frac{1}{2}]$, $g_1(x_0)=g(x_0)\in O'$.
So, $x_0\in g_1^{-1}(O')$.
So, $g^{-1}([0,1]\cap O')\subset [0,\frac{1}{2}]\cap g_1^{-1}(O')$.

Let $x_0\in[0,\frac{1}{2}]\cap g_1^{-1}(O')$.
Then, $g(x_0)\in [0,1]$ since the codomain of $g$ is $[0,1]$.
Since $g(x)=g_1(x)$ on $[0,\frac{1}{2}]$, $g(x_0)=g_1(x_0)\in O'$.
So, $x_0\in g^{-1}([0,1]\cap O')$.

Since $g_1^{-1}(O')$ is an open subset of $\mathbb{R}$, $g^{-1}([0,1]\cap O')=[0,\frac{1}{2}]\cap g_1^{-1}(O')$ is open in the topological subspace of $[0,\frac{1}{2}]$ of $\mathbb{R}$.

I guess my solution is unnecessarily long.
Please give me a better solution.

Answer 1

Yes, your solution is unnecessarily long. A shorter solution is, for example, this: Let $g:[0,\frac{1}{2}]\to [0,1]$ be the map defined by $g(x)=2x$. The map $g$ is obviously continuous. Since $f_1=f\circ g$, and since $f$ is continuous by hypothesis, it follows that the composition $f_1=f\circ g$ is also continuous.