Let $(X,x_0)$ and and $(Y,y_0)$ be compactly generated well-pointed Hausdorff-spaces. Then the smash product

$$X \wedge Y:= X \times Y/ (\{x_0\} \times Y \cup X \times \{y_0\})$$

is supposed to be a compactly compactly generated well-pointed Hausdorff-space as well. What I don't quite see is the Hausdorff part.To be precise, I don't see how can choose an open neigbourhood for $[(x_0,y_0)]$.

Thank you.

  • $\begingroup$ What do you mean by "well-pointed" ? $\endgroup$
    – Jean Marie
    Mar 12, 2017 at 7:52
  • $\begingroup$ The inclusion of the base point is a cofibration. $\endgroup$
    – Joe91
    Mar 12, 2017 at 7:53
  • $\begingroup$ Thanks for the answer but it is even darker now maybe because it's algebraic topology and that I am only acquainted with classical topology... sorry for my incompetence. $\endgroup$
    – Jean Marie
    Mar 12, 2017 at 7:59

2 Answers 2


This is routine just using the definition of the smash product and the Hausdorff property.

We need to find disjoint open neighbourhoods of $[(x_0,y_0)]$ and $[(x,y)]$ in $X\wedge Y$. We know $(x,y)$ is not in $\{x_0\} \times Y \cup X \times \{y_0\}$, so $x\neq x_0$ and $y\neq y_0$. Using the Hausdorff property we get:

  • disjoint open neighbourhoods $U_0, U\subseteq X$ such that $x_0\in U_0$ and $x\in U$, and
  • disjoint open neighbourhoods $V_0, V\subseteq Y$ such that $y_0\in V_0$ and $y\in V$.

Define $W_0=(U_0\times Y) \cup (X \times V_0)$ and $W=U\times V$ as subsets of $X\times Y$. Note that these are disjoint open neighbourhoods with $(x_0,y_0)\in W_0$ and $(x,y)\in W$. These sets also respect the quotient map so give disjoint open neighbourhoods of $[(x_0,y_0)]$ and $[(x,y)]$ in $X\wedge Y$ as required.


Hausdorff property: Since $\{x_0\}\times Y\cup X\times \{y_0\}$ is a closed subset of $X\times Y$, your question reduces to be : If $Z$ is Hausforff, A $\subset$ Z is closed, then show that the quotient space $Z/A$ is Hausdorff, note that the quotient map is an open map under the quotient topology.

  • $\begingroup$ It's not true in general that collapsing a closed subset of a Hausdorff space gives a Hausdorff quotient; otherwise every Hausdorff space would be normal - see for example math.stackexchange.com/a/1744671/3643 $\endgroup$ Jun 25, 2017 at 9:26
  • $\begingroup$ @ColinMcQuillan Thank you. $\endgroup$
    – LipCaty
    Jun 30, 2017 at 6:34

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