Equivalence of two topologies on the topological join In his famous paper on universal bundles, Milnor defined the strong topology on the join $X\ast Y$ of two spaces $X$ and $Y$ as the initial topology induced by the "coordinate" functions.
There is also a quotient topology on the join, which he called weak. The whole idea was that the diagonal action of a topological group $G$ on its multi-join $G\ast\ldots\ast G$ is continuous in the strong topology  (I think it is not known if it is true for the weak topology). The two topologies coincide when $X$ and $Y$ are compact Hausdorff. A proof of this result can be found in Ronald Brown's book Topology and Groupoids in the section about joins.
Question: Do we know some other instances when the strong and weak topologies coincide for the join? For example, when $X$ and $Y$ are locally compact Hausdorff or more generally compactly generated Hausdorff.
 A: The join $X\ast Y$ is the quotient of $X\times Y\times I$ by the relations
$$(x,y,0)\sim (x',y,0),\qquad\qquad (x,y,1)\sim (x,y',1).$$
The Milnor join $X\overline\ast Y$ has the same underlying set as $X\ast Y$, but is equipped with the initial topology determined by the three obvious 'coordinate' maps
$$t:X\overline\ast Y\rightarrow I,\qquad x:t^{-1}[0,1)\rightarrow X,\qquad y:t^{-1}(0,1]\rightarrow Y.$$
The identity map
$$\varphi:X\ast Y\rightarrow X\overline\ast Y$$
is a continuous bijection. This expresses the fact that (with the usual conventions), the Milnor topology on $X\overline\ast Y$ is weaker than the join topology on $X\ast Y$ (ie. $X\overline\ast Y$ has fewer open sets than $X\ast Y$).
The fact that the Milnor topology is weaker than the join topology is useful. For instance
$$(X\overline \ast Y)\overline\ast Z\cong X\overline \ast (Y\overline\ast Z)$$
for any spaces $X,Y,Z$, while the join operation may fail to be associative.
If $X,Y$ are Hausdorff, then it is easily checked that so is $X\overline\ast Y$, which means that so too is $X\ast Y$. On the other hand, if $X,Y$ are compact, then so is $X\ast Y$, and hence also $X\overline\ast Y$. By the minimality of compact Hausdorff topologies we thus have the following.

$\varphi$ is a homeomorphism when $X,Y$ are both compact $T_2$.

In general $\varphi$ can fail to be a homeomorphism. Nevertheless it has some useful properties.

$\varphi$ is always a homotopy equivalence.

Proof: A homotopy inverse to $\varphi$ is given by the map
$$[x,y,t]\mapsto \begin{cases} [x,y,0]&0\leq t\leq \frac{1}{3}\\
[x,y,3t-1]&\frac{1}{3}\leq t\leq \frac{2}{3}\\
[x,y,1]&\frac{2}{3}\leq t\leq 1. \end{cases}\qquad\square$$
Really we would like to say something more general about when $\varphi$ is a homeomorphism. To do this we will compare the joins with the pushout
$$X\times CY\cup CX\times Y.$$
Here
$$CZ=Z\times I/Z\times\{1\}$$
denotes the cone on $Z$. We have $Z\subseteq CZ$ embedded as a closed subspace at level $t=0$.
Note that because $X\times Y$ is closed in both $CX\times Y$ and $X\times CY$, the pushout topology on $X\times CY\cup CX\times Y$ agrees with the product topology it inherits from $CX\times CY$.
Define
$$\alpha:X\overline\ast Y\rightarrow CX\times Y\cup X\times CY,\qquad
[x,y,t]\mapsto \begin{cases}([x,0],[y,1-2t])&0\leq t\leq \frac{1}{2}\\
([x,1-2(1-t)],[y,0])&\frac{1}{2}\leq t\leq 1.\end{cases}$$
The continuity of this map is easily checked using the previous observation: it suffices to consider its composites with the projections to $CX,CY$.

The map $\alpha:X\overline\ast Y\rightarrow CX\times Y\cup X\times CY$ is a homeomorphism.

Proof: Indeed, an inverse is given by
$$\beta:X\times CY\cup CX\times Y\rightarrow X\overline\ast Y,\qquad \begin{cases}([x,0],[y,t])\mapsto[x,y,\frac{1}{2}(1-t)],\\([x,s],[y,0])\mapsto[x,y,1-\frac{1}{2}(1-s)].\end{cases}$$
To check its continuity use the pushout topology on $X\times CY\cup CX\times Y$ and study the composites of $\beta$ with the coordinate functions on $X\overline\ast Y$. $\square$
Composing $\alpha$ with $\varphi$ yields a continuous bijection, which by abuse we will continue to denote
$$\alpha:X\ast Y\rightarrow CX\times Y\cup X\times CY.$$

If both $X,Y$ are locally compact, then $\alpha:X\ast Y\rightarrow CX\times Y\cup X\times CY$ is a homeomorphism.

For clarity, local compactness of $Z$ means that each of its points has a local base of compact neighbourhoods.
Proof: The inverse is given by the map $\beta$ defined above. We must demonstrate its continuity for the new domain.
We use the pushout topology on $CX\times Y\cup X\times CY$, meaning that this space is a quotient of $CX\times Y\sqcup X\times CY$. The assumptions that $X,Y$ are locally compact means that the each of the two maps
$$X\times Y\times I\rightarrow X\times CY,\qquad X\times I\times Y\rightarrow CX\times Y$$
is a quotient. Thus we have a quotient map
$$q:X\times Y\times I\sqcup X\times I\times Y\rightarrow CX\times Y\cup X\times CY$$
Now continuity of $\beta$ is equivalent to the continuity of the two composites
$$\beta_1:X\times Y\times I\xrightarrow{q_1} X\times CY\cup CX\times Y\xrightarrow\beta X\ast Y$$
$$\beta_2:X\times I\times Y\xrightarrow{q_2} X\times CY\cup CX\times Y\xrightarrow\beta X\ast Y$$
where $q_1,q_2$ are the two components of $q$.
Clearly $\beta_1$ factors as a continuous map $X\times Y\times I\rightarrow X\times Y\times I$ followed by the canonical map $X\times Y\times I\rightarrow X\ast Y$. Simlarly $\beta_2$ factors as a composite of a continuous map $X\times I\times Y\rightarrow X\times Y\times I$ and the quotient to $X\ast Y$. Thus we may conclude that $\beta$ is continuous. $\square$.
As a corollary we obtain the following.

If both $X,Y$ are locally compact, then $\varphi:X\ast Y\rightarrow X\overline\ast Y$ is a homeomorphism. $\square$

The local compactness of both spaces is essential for this result. As an example consider
$$X=\mathbb{R}\setminus\{\frac{1}{n}\mid n\in\mathbb{N}\}.$$
The product topology on $X\times C\mathbb{N}$ does not agree with the quotient topology inherited from the map $X\times \mathbb{N}\times I\rightarrow X\times C\mathbb{N}$. Consequently the join topology on $X\ast \mathbb{N}$ is strictly finer than the Milnor topology on $X\overline\ast \mathbb{N}$.
Note that $\mathbb{N}$ is locally compact, so the failure of the continuous bijection $\varphi$ to be a homeomorphism is attributed soley to $X$ (which is not locally closed in $\mathbb{R}$, and hence not locally compact).
Finally note that both $X$ and $\mathbb{N}$ are first-countable and Hausdorff, so therefore both compactly generated. Thus compact generation is not sufficient to make $\varphi$ a homeomorphism.
Of course, if we work in the category $CGTop$ of compactly generated spaces (and in particular use its product), then $X\times_kC\mathbb{N}$ is a quotient of $X\times \mathbb{N}\times I$.

If $X,Y$ are compactly generated spaces, then $k(X\ast Y)\rightarrow k(X\overline\ast Y)$ is a homeomorphism.

The proof is as above: both spaces have the same topology as the pushout $X\times_kCY\cup CX\times_kY$ when computed in $CGTop$.
