Why is the Koch curve homeomorphic to $[0,1]$? Henning Makholm has provided a nice proof that the limiting curve is a continuous function from $[0,1]$ to the plane. I was curios if the function is homeomorphism. A quick search gave me many sources mentioning that $[0,1]$ and the Koch curve are indeed homeomorphic (as a proof that they have the same topological dimension), yet without showing why are they homeomorphic. What would be the simplest way to show it?

Yes, the standard parameterization of the snowflake curve is injective.
To see this easily, instead of making each approximant a curve, we can make it a chain of linked half-open equilateral triangles:

Each of the red triangles includes the side between the entry and exit point, but doesn't the two other sides. In each step each triangle splits into four smaller triangles that are completely contained in the original one. Therefore, at each step all of the small triangles are disjoint, except for the shared corner points where the curve moves from one triangle to the next (black dots in the diagram).
Now if we have a parameter value $x\in[0,1]$ and want to find $\gamma(x)$, write $x$ as a 4-ary fraction and then each digit will decide one of the 4 subtriangles to move into at each step in the construction. $\gamma(x)$ is then the unique point that is in the intersection of a sequence of smaller and smaller nested triangles.
If $x$ and $y$ are different parameter values, then their 4-ary expansions must differ from some point. But that means that $\gamma(x)$ and $\gamma(y)$ are in different small triangles at that step in the construction. Since the triangles don't have points in common, $\gamma(x)$ and $\gamma(y)$ are not the same point.
