Some quotient of $[0,1]$ is homeomorphic to $[0,1] \times [0,1]$

Question

I have been unable to solve this problem. I imagine you would start with some bijection $$f:[0,1] \rightarrow [0,1] \times [0,1]$$ But I don't get how any topological argument could be made here with an arbitrary bijection.

Answer 1

The answer is pretty easy using Mike Miller's hint. Let $f:[0,1] \rightarrow [0,1] \times [0,1]$ be a continuous surjection, and set $x \sim y$ if and only if $f(x) = f(y)$. If we let $X = [0,1]$ the quotient map, and $p: [0,1] \rightarrow X/ \sim$, then there is a unique continuous map $\bar{f}: X \rightarrow [0,1] \times [0,1]$ for which $\bar{f} \circ p = f$.

Since $\bar{f}$ is a continuous bijection, it's automatically a homeomorphism ($X$ is compact and $[0,1] \times [0,1]$ is Hausdorff).