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This must be an elementary question, as I couldn't find any proofs on the Internet, but I still can't do it. And yes, Hatcher says that the join is not actually associative for general topological spaces, but it is for locally Hausdorff ones. My definition of the join is a quotient of a product.

My problem is that I don't really understand the topology of the join. Is there a basis? What do open sets look like? I think I can find a bijection that is supposed to be a homeomorphism between $(A*B)*C$ and $(B*C)*A$ (this should suffice, as commutativity seems easy): $$[[a,b,t],c,u] -> [[b,c,\frac{u}{1-(1-t)(1-u)}],a,(1-t)(1-u)].$$

Function in the third coordinate is not continuous at (0,0), so we can't just define it in the product and then push it down to the quotient. At (0,0) we can define it to be anything(e.g. $\frac{1}{2}$) as the 1 in the last coordinate collapses the first space ($B*C$) anyway.

The context of this question: I am trying to prove that $S^p * S^q = S^{p+q+1}$ and my idea was first to prove associativity and then just expand everything as multiple joins of $S^0$. There may be easier ways to prove the fact about spheres, but after having been stuck on this I really want to see the proof of associativity. The question about spheres is from Rolfsen's Knots and Links.

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Maybe you could check that both joins are homeomorphic to the space of formal linear combinations $t_1a + t_2b + t_3c$ where $t_1 + t_2 + t_3 = 1$, topologized as a subset of $A\times B\times C\times\Delta^2$ (at least when the topologies on $A$, $B$ and $C$ are sufficiently nice)? – Mark Grant Oct 29 '12 at 9:39
@Mark: $A\times B\times C\times \Delta^2$ is a good way to look at it. But you want a subset, not a quotient. – Tom Goodwillie Oct 29 '12 at 14:15
I mean a quotient, not a subset! – Tom Goodwillie Oct 29 '12 at 14:15
@Tom: Oh yes, you're right. – Mark Grant Nov 1 '12 at 15:15
up vote 7 down vote accepted

Actually there are two useful topologies on the join $X * Y$ using initial topologies, or final topologies. The former is useful for maps into the join, and the latter for maps out of the join. With initial topology, the join is associative. With final topologies, the standard problem is that the product of identification maps is not an identification map, unless one works in a convenient category of spaces. These questions are discussed in Section 5.7 of Topology and Groupoids, where associativity with the initial topology is proved. (This was in the 1968, 1988 editions!) Of course the two topologies agree on compact Hausdorff spaces, but I have not managed to write down a proof that the two topologies are equivalent in the category of compactly generated spaces. Am I missing something easy?

Later: just to add some details. One thinks of the join as lines joining points of $X$ to points of $Y$ in some Euclidean space (see Tom's answer). So one writes these elements as formal combinations $rx+sy$ where $x \in X, y \in Y$ and $0 \leqslant r,s$ and $ r+s =1$. Of course $0x+1y$ means just $y$ and $1x + 0y$ means just $x$. The initial topology is with respect to the functions $$rx+sy \mapsto r, rx+sy \mapsto s, rx+sy \mapsto x, rx+sy \mapsto y$$ which are well defined on subsets of $X*Y$. With this notation, associativity becomes easy, since a point of $X*Y*Z$ is of the form $rx+sy +tz$, with the obvious conventions and conditions; continuity of the associativity map rolls out with these initial topologies.

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If $A$ and $B$ are subsets of $\mathbb R^n$ then you can map $A\times I\times B$ to $\mathbb R^n$ by $(a,t,b)\mapsto (1-t)a+tb$. If the resulting continuous map $A*B\to \mathbb R^n$ happens to be one to one then you can ask whether it gives a homeomorphism to its image. If $A$ and $B$ are compact, then the answer is necessarily yes, since a continuous bijection from a compact space to a Hausdorff space is always a homeomorphism.

This suffices to see that the join of spheres is a sphere.

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Thanks! That does answer the question about spheres. – shestipalov Oct 29 '12 at 22:01

You may think of join $A*B$ as a sphere in the product $(\mathop{\rm Cone}A)\times (\mathop{\rm Cone}B)$.

Once you made all this precise in topological category, the associativity will follow from associativity of Cartesian product.

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