I've been looking at a free topology PDF that I downloaded somewhere and the proof for this theorem is bothering me because it skips some crucial steps. I have a couple of questions. First, I'll show you the proof.
Theorem Let $X$ be a Hausdorff topological space and let $f: X \to Y$ be a homeomorphism for some other topological space $Y$. Then $Y$ is also Hausdorff.
Proof
Suppose we have two homeomorphic spaces $X$ and $Y$. Let $X$ be Hausdorff.
Pick two distinct points $x_1, x_2 \in X.$
Plugging this into our homeomorphism, we have $f(x_1) = y_1$ and $f(x_2) = y_2$.
Since $f$ is one-to-one, $y_1 = y_2 \iff x_1 = x_2$. But we said $x_1 \neq x_2$. Therefore, $y_1 \neq y_2.$
Take two open sets around $x_1$ and $x_2, O_1$ and $O_2,$ such that $O_1 \bigcap O_2 = \emptyset.$
Consider $f(O_1)$ and $f(O_2)$. These sets are open in $Y$ with $y_1 \in f(O_1)$ and $y_2 \in f(O_2)$ by our homeomorphism.
Finally, because $f$ is an onto function, $f(O_1) \bigcap f(O_2) = \emptyset.$
Therefore, $Y$ is Hausdorff. $\hspace{1cm} \square$
First question: How do we know that disjoint $O_1$ and $O_2$ exist in our topology?
Second question: How does $f$ being surjective imply that $f(O_1) \bigcap f(O_2) = \emptyset$? All that surjectivity tells us is that $f^{-1}(f(O_i))$ exists.
Best Answer
I think a correct proof would go like this:
Pick some $y_1 \neq y_2 \in Y$. Since $f$ is surjective, there exists $x_1\neq x_2 \in X$ such that $f(x_1)=y_1$ and $f(x_2)=y_2$. The space $X$ is Hausdorff, so there exists disjoint open neighborhoods $O_1, O_2$ of $x_1$ and $x_2$ respectively. Since $f$ is injective, $f(O_1) \cap f(O_2) = \emptyset$. Also, since $f$ is an open map (this is equivalent to the fact that $f^{-1}$ is continuous), $f(O_1)$ and $f(O_2)$ are open neighborhoods of $y_1$ and $y_2$. Thus, $Y$ is indeed Hausdorff.