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Let $X$ be a topological space. We denote by $\tau_\mathbb{R}$ the final topology induced by the family of continuous maps $\varphi:\mathbb{R}\rightarrow X$, and by $\tau_I$ the final topology induced by the family of continuous maps $\psi:I\rightarrow X$, where $I$ is the closed unit interval $[0,1]$.

I want to show that $\tau_I=\tau_\mathbb{R}$.

The proof that $\tau_\mathbb{R}\subset \tau_I$ isn't too difficult:

Suppose $U$ is an element of $\tau_\mathbb{R}$. We wish to show that for every continuous map $\varphi:I\rightarrow X$ we have $\varphi^{-1}(U)$ is open. To achieve this, we extend $\varphi:I\rightarrow X$ to a map $\overline{\varphi}:\mathbb{R}\rightarrow X$ by defining $$\overline{\varphi}(x)=\begin{cases} \varphi(x) & x\in I \\ \varphi(1) & x>1 \\ \varphi(0) & x<0 \end{cases}$$ which is clearly continuous. Thus, since $U$ is an element of $\tau_\mathbb{R}$, we have $\overline{\varphi}^{-1}(U)$ is open, by which $I\cap \overline{\varphi}^{-1}(U)=\varphi^{-1}(U)$ is open in $I$.

However, I can't think of any good approach to take in showing that $\tau_I\subset \tau_\mathbb{R}$. Any help would be greatly appreciated.

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Let $f:\mathbb{R}\to X$ be continuous, and take $U\in\tau_I$.

To show that $f^{-1}(U)$ is open, it is sufficient to show that $f^{-1}(U)\cap J$ is open in $J$ for every closed interval $J\subset\mathbb{R}$.

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