Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to prove that if in (Haus, U) a morphism $f: X \rightarrow Y$ is not dense, it isn't an epimorphism. To prove this I need the following construction:

Suppose $M = \overline{f(X)}$ is closed in $Y$. We can form the coproduct $j_{1}, j_{2}: Y \rightarrow Y_{1} + Y_{2}$ of two copies of $Y$ and then identify the corresponding points in the copies of $M$ by $\varphi : Y_{1} + Y_{2} \rightarrow Z$, where $Z$ carries the final topology.

Now i can conclude that $(\varphi \circ j_{1}) \circ f = (\varphi \circ j_{2}) \circ f,$ but $\varphi \circ j_{1} \neq \varphi \circ j_{2}$.

The only problem is that to finish the proof, I need to show that the space $Z$ is a Hausdorff space.

Can anyone explain me how to do this? I think I need to work out the different options (trying to separate a point in $M$ and in $Y_{1}$, to points in $M$, etc.), but I don't know how to start my proof.

share|improve this question
    
Hint: if your two points in $Z$ correspond to different points in $Y$, this is easy from the Hausdorffness of $Y$. The only difficulty is if they correspond to the same point in $Y$. –  Chris Eagle Mar 19 '12 at 14:34

1 Answer 1

up vote 4 down vote accepted

Let $z_1,z_2$ be distinct points of $Z$. Let $M_1$ and $M_2$ be the copies of $M$ in $Y_1$ and $Y_2$, respectively. Now consider the following possibilities.

  1. There is an $i\in\{1,2\}$ and there are $y_1,y_2\in Y_i$ such that $\varphi(y_1)=z_1$ and $\varphi(y_2)=z_2$. Use the fact that $Y_i$ is Hausdorff to get a separation of $z_1$ and $z_2$.

  2. Not case (1). There still must be $y_1,y_2\in Y_1+Y_2$ such that $\varphi(y_1)=z_1$ and $\varphi(y_2)=z_2$, but they cannot both be in the same $Y_i$. Clearly we may assume that $y_1\in Y_1$ and $y_2\in Y_2$, but we can go further than this: we may assume that $y_1\in Y_1\setminus M_1$ and $y_2\in Y_2\setminus M_2$. (Why?) Now use the fact that $M_i$ is closed in $Y_i$ for $i=1,2$ to get your separation of $z_1$ and $z_2$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.