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Help me please to solve this problem. Let's $f : X \to Y$ be one-to-one and onto, prove that:

a) If $f : X\to Y $ a continuous function and $Y$ is Hausdorff then $X$ is Hausdorff.

b) If $f$ is open or closed and $X$ is Hausdorff then $Y$ is Hausdorff.


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  1. Since $f$ is continuous, $f^{-1}[U]$ is open in $X$ whenever $U$ is open in $Y$.

  2. The case when $f$ is open is almost exactly like the first problem. For the other case, show that if $f$ is closed, then $f$ is also open.

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The best advice I can give you is to write down the definitions of everything and the results should follow immediately.

  1. $X$ is Hausdorff if for every $x,y\in X$ distinct there are $U,V$ open such that $x\in U, y\in V$ and $U\cap V=\varnothing$.

  2. $f\colon X\to Y$ is continuous if for every $U$ open in $Y$ the preimage, $f^{-1}[U]=\{x\in X\mid f(x)\in U\}$ is open in $X$.

Now we want to prove that if we have a continuous bijection from $X$ to $Y$ and $Y$ is Hausdorff then $X$ is Hausdorff. Suppose, if so, that $a,b\in X$ are two distinct points.

Since $f$ is a bijection $f(a)\neq f(b)$. Therefore there are $U,V$ open in $Y$ which are disjoint and $f(a)\in U$ and $f(b)\in V$. By the continuity of $f$ we have that $f^{-1}[U]$ and $f^{-1}[V]$ are open, and disjoint. Furthermore, $a\in f^{-1}[U]$ and $b\in f^{-1}[V]$ (since $f(a)\in U$ and $f(b)\in V$). Therefore $X$ is Hausdorff (we found the disjoint open sets).

For the second question, note that if $f$ is a bijection which is either open or closed; then its inverse is a continuous bijection (why?), so we reduce the second question to the first by taking $f^{-1}$ as our function, and now the range is Hausdorff.

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