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The answer to 1 is yes. For the purpose of this answer, a bipointed space is a topological space $J$ equipped with distinct closed points $e_0$ and $e_1$. As you say, for any bipointed space $J = (J, e_0, e_1)$, we can form a new bipointed space $J \vee J$ by taking the disjoint union of two copies of $J$, identifying the first $e_1$ with the second $e_0$, and giving the resulting space the obvious pair of basepoints.

Theorem: In the category of bipointed spaces $J$ equipped with a map $J \to J \vee J$, the initial terminal object is the bipointed space $([0, 1], 0, 1)$ equipped with the map "multiplication by 2" from $[0, 1]$ to $[0, 1] \vee [0, 1] \cong [0, 2]$.

Or informally: $[0, 1]$ has the structure needed in order to be able to define and compose paths, and is universal as such.

The theorem is proved here, and is a variant of a result of Peter Freyd's (which characterized the interval set-theoretically and order-theoretically, but not topologically). The idea that $[0, 1]$ is universal with the structure needed for homotopy theory is expanded on in these talk slides.

show/hide this revision's text 1

The answer to 1 is yes. For the purpose of this answer, a bipointed space is a topological space $J$ equipped with distinct closed points $e_0$ and $e_1$. As you say, for any bipointed space $J = (J, e_0, e_1)$, we can form a new bipointed space $J \vee J$ by taking the disjoint union of two copies of $J$, identifying the first $e_1$ with the second $e_0$, and giving the resulting space the obvious pair of basepoints.

Theorem: In the category of bipointed spaces $J$ equipped with a map $J \to J \vee J$, the initial object is the bipointed space $([0, 1], 0, 1)$ equipped with the map "multiplication by 2" from $[0, 1]$ to $[0, 1] \vee [0, 1] \cong [0, 2]$.

Or informally: $[0, 1]$ has the structure needed in order to be able to define and compose paths, and is universal as such.

The theorem is proved here, and is a variant of a result of Peter Freyd's (which characterized the interval set-theoretically and order-theoretically, but not topologically). The idea that $[0, 1]$ is universal with the structure needed for homotopy theory is expanded on in these talk slides.