# Proving a space is Hausdorff, given a specific construction

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.

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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

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 $z_2$.