Hi.
A fast question (i dont want a proof):
Let and two topological spaces. Itīs not Hausdorff.
Let X_1, T_1)\to(X_2, T_2)" alt="fX_1, T_1)\to(X_2, T_2)" /> a function continuous. Its possible that 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.
Take an example that is an indiscrete topology given on and is a Hausdorff space such that f is a constant map. We see that f is continuous.
Not necessarily true. Consider is any non-Hausdorff topological space and is a discrete topology (which is Hausdorff) given on . All maps having as a domain and as a codomain are continuos (You can easily check this by the definition of a continuous function).
Consider this counterexample.
Let be the real line with the standard topology; let be the product topological space , where is the with the indiscrete topology. Then is not a Hausdorff space but is a Hausdorff space. You can construct a surjective continuous map given , where and .
The counterexample I showed you still works even if the continuous function in the example is surjective.
Consider an identity map between with the discrete topology and 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.