Let $X$ be Hausdorff space and $f$ is a continuous function from $[0,1]$ to $X$. If $f$ is one-one, then image of $f$ is homeomorphic to $[0,1].$
I did something like defining mapping $g$ from image of $f$ to $[0,1]$ as $g(x)=y$ where $f(y)=x.$ Everything is going fine except $g$ is continuous means I don't know how to show continuity of $g$. Any hint please. Thank you.
To show that $g$ is continuous, you have to show that the preimage of a closed set under $g$ is closed. By definition of $g$ this is equivalent to the fact that $f$ maps closed sets to closed sets.
Now assume that $M \subset [0,1]$ is closed. Then it is compact and by continuity of $f$ you get that $f(M)$ is also compact. Now $X$ is a Hausdorff-space which means that compact subsets of $X$ are closed and therefore $f(M)$ is closed.
Lemma: The continuous image of a compact space is compact.
Little Lemma: A compact subset of a Hausdorff space is a closed set.
Littler Lemma: A subspace $Z$ of a Hausdorff space $X$ is also a Hausdorff space.
Littlest Lemma: If $f:Y\to X$ is continuous and $f(Y)$ has the subspace toology as a subspace of $X$ then $f:Y\to f(Y)$ is continuous.
Theorem: If $Y,Z$ are compact Hausdorff spaces and $f:Y\to Z$ is a continuous bijection then $f$ is a homeomorphism.
Apply to $Y=[0,1]$ and $Z=f(Y)\subset X,$ noting that $[0,1]$ is compact Hausdorff.
