1. ## Topology

Hi.

A fast question (i dont want a proof):

Let $(X_1, T_1)$ and $(X_2, T_2)$ two topological spaces. $(X_1, T_1)$ It´s not Hausdorff.
Let $fX_1, T_1)\to(X_2, T_2)" alt="fX_1, T_1)\to(X_2, T_2)" /> a function continuous. Its possible that $(X_2, T_2)$ to be Hausdorff?

I think no.

Thanks

2. Originally Posted by asub1
Hi.

A fast question (i dont want a proof):

Let $(X_1, T_1)$ and $(X_2, T_2)$ two topological spaces. $(X_1, T_1)$ It´s not Hausdorff.
Let $fX_1, T_1)\to(X_2, T_2)" alt="fX_1, T_1)\to(X_2, T_2)" /> a function continuous. Its possible that $(X_2, T_2)$ to be Hausdorff?

I think no.

Thanks
Not true. Take the line with the double origin, and map it to the real line as the identity and send the second origin also to 0. The map is continuous and the reals are Hausdorff.

It is true however if you swap the spaces around i.e. f cts and X_2 not Hausdorff implies that X_1 is not Hausdorff.

3. Originally Posted by asub1
Hi.

A fast question (i dont want a proof):

Let $(X_1, T_1)$ and $(X_2, T_2)$ two topological spaces. $(X_1, T_1)$ It´s not Hausdorff.
Let $fX_1, T_1)\to(X_2, T_2)" alt="fX_1, T_1)\to(X_2, T_2)" /> a function continuous. Its possible that $(X_2, T_2)$ to be Hausdorff?

I think no.

Thanks
Take an example that $T_1$ is an indiscrete topology given on $X_1$ and $(X_2, T_2)$ is a Hausdorff space such that f is a constant map. We see that f is continuous.

Originally Posted by Focus
It is true however if you swap the spaces around i.e. f cts and X_2 not Hausdorff implies that X_1 is not Hausdorff.
Not necessarily true. Consider $(X_2, T_2)$ is any non-Hausdorff topological space and $T_1$ is a discrete topology (which is Hausdorff) given on $X_1$. All maps having $(X_1, T_1)$ as a domain and $(X_2, T_2)$ as a codomain are continuos (You can easily check this by the definition of a continuous function).

4. Hi!

I understand your example aliceinwonderland, but ...

$(X_1, T_1)$ not Hausdorff, $pX_1, T_1) \to (X_2, T_2)" alt="pX_1, T_1) \to (X_2, T_2)" /> continuous and surjective, implies $(X_2, T_2)$ not Hausdorff.

Thanks.

5. Originally Posted by asub1
Hi!

I understand your example aliceinwonderland, but ...

$(X_1, T_1)$ not Hausdorff, $pX_1, T_1) \to (X_2, T_2)" alt="pX_1, T_1) \to (X_2, T_2)" /> continuous and surjective, implies $(X_2, T_2)$ not Hausdorff.

Thanks.
Consider this counterexample.

Let $X_2$ be the real line $\mathbb{Re}$ with the standard topology; let $X_1$ be the product topological space $X_2 \times Y_1$, where $Y_1$ is the $\mathbb{Re}$ with the indiscrete topology. Then $X_1$ is not a Hausdorff space but $X_2$ is a Hausdorff space. You can construct a surjective continuous map $f:X_2 \times Y_1 \rightarrow X_2$ given $f(x_2, y_1) = x_2$, where $x_2 \in X_2$ and $y_1 \in Y_1$.

6. Originally Posted by aliceinwonderland
Not necessarily true. Consider $(X_2, T_2)$ is any non-Hausdorff topological space and $T_1$ is a discrete topology (which is Hausdorff) given on $X_1$. All maps having $(X_1, T_1)$ as a domain and $(X_2, T_2)$ as a codomain are continuos (You can easily check this by the definition of a continuous function).
Yes sorry, you need the function to be surjective.

7. Originally Posted by Focus
Yes sorry, you need the function to be surjective.
The counterexample I showed you still works even if the continuous function in the example is surjective.
Consider an identity map between $\mathbb{Re}$ with the discrete topology and $\mathbb{Re}$ with the indiscrete or cofinite topology. Then the map is continuous and surjective, but the domain of the map is Hausdorff and the codomain of the map is non-Hausdorff.