This is my second question from Hatcher chapter 0 (and final I think). For $X$, $Y$ CW complexes, it asks one to show that $$X \ast Y = S(X \wedge Y)$$ by showing $$X \ast Y/(X \ast y_0 \cup x_0 \ast Y) = S(X \wedge Y) / S(x_0 \wedge y_0),$$ where $\ast$ is topological join, $\wedge$ is smash product, $S$ is suspension and $=$ is homeomorphism.

Unfortunately I am quite at loss on this question besides from knowing some definitions of these spaces. $S(x_0\wedge y_0)$ is suspension of a point which is an interval I. $X \wedge Y$ is the space obtained by the quotient $X \times Y / X \vee Y$ where $X \vee Y$ is wedge sum of $X$ and $Y$ at the point $x_0$ and $y_0$. Suspension of this is its product with interval $I$ and collapsing the end points. Overall we have $$X \times Y \rightarrow X \times Y / X \vee Y \rightarrow (X \times Y / X \vee Y) \times I \rightarrow Z = (X \times Y / X \vee Y) \times I/(\text{contract $t=0$ and $t=1$ to a point}) \rightarrow Z / S(x_0\wedge y_0).$$

$(X \ast {y_0} \cup {x_0} \ast Y)$ is the union of two cones over $X$ and $Y$ and finally $X \ast Y$ could be seen as union of cones $C(X,y)$ for all $y$ or the other way around. Then this space is $$X \times Y \times I \rightarrow A= X \times Y \times I/(\text{contract $Y$ at $t=0$ and $X$ at $t=1$}) \rightarrow A/(X \ast {y_0} \cup {x_0} \ast Y).$$

However I cant see how to relate these operations. I know some theorems on these space making homotopy relations but this questions requires this to be a homeomorphism.

Thanks a lot.

  • $\begingroup$ See proposition 4I.1 $\endgroup$
    – user641
    Commented Aug 1, 2012 at 22:57
  • $\begingroup$ Recall that Hatcher's symbol for homotopy equivalence is given by an underlined tilde- in my printing, this is the symbol that appears in the question. I believe that the question is only asking about homotopy equivalence. $\endgroup$
    – KReiser
    Commented Aug 1, 2012 at 23:01
  • $\begingroup$ Yes, I agree with KReiser that you are trying to show $X*Y\simeq S(X\wedge Y)$ where $\simeq $ is homotopy equivalence, and you are trying to do so by showing that the quotients are homeomorphic. So this breaks the problem into two separate parts. For the first part you view both spaces as quotients of $X\times Y \times I$ (just write out carefully what points you are identifying), but for the second, you need to use proposition 0.17 to relate the original spaces with the quotients. $\endgroup$
    – Aaron
    Commented Aug 1, 2012 at 23:13

2 Answers 2


Let's show the reduced versions are homeomorphic, which will show the originals are homotopic (they are not equal in general).

The join can be thought of as "lines" from $X$ to $Y$, with some collapsing. The relations are: $$ (x_1,y,0)\sim (x_2,y,0);$$ $$ (x,y_1,1)\sim (x,y_2,1).$$

The reduced version also collapses $x_0\ast Y$ and $X\ast y_0$. So the additional relations are: $$ (x_0,y,t)\sim (x_0,y_0,0);$$ $$ (x,y_0,t)\sim (x_0,y_0,0).$$

We can derive further relations too: $$(x,y,0)\sim(x_0,y,0)\sim(x_0,y_0,0);$$ $$(x,y,1)\sim (x,y_0,1)\sim(x_0,y_0,0).$$

The smash product is gotten from $X\times Y$ by collapsing $X\times y_0$ and $x_0\times Y$. The suspension of that can be thought of as $X\times Y\times I$, with the relations: $$ (x,y_0,t)\sim (x_0,y_0,t);$$ $$ (x_0,y,t)\sim (x_0,y_0,t);$$ $$ (x,y,1)\sim (x_0,y_0,1);$$ $$ (x,y,0)\sim (x_0,y_0,0).$$

The reduced suspension adds the relation $$ (x_0,y_0,t)\sim (x_0,y_0,0).$$

Now it is not hard to see you are quotienting out by the same relations for both constructions. Namely,

$$ (x,y_0,t)\sim (x_0,y_0,0);$$ $$ (x_0,y,t)\sim (x_0,y_0,0);$$ $$ (x,y,0)\sim (x_0,y_0,0);$$ $$ (x,y,1)\sim (x_0,y_0,0).$$


Some pictures from "Topology and Groupoids", Chapter 5, which you may find helpful are the join $X * Y$ as the join

and the subspace to be collapsed to a point to give the suspension of the smash product is subspace where the two vertices on the mid line are the base points.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .