[proofplan] We realize the join as the quotient of $S^a \times S^b \times [0,1]$ and construct an explicit map to the unit sphere in $\mathbb{R}^{a+1} \times \mathbb{R}^{b+1}$ by weighting the two sphere coordinates with $\sqrt{1-t}$ and $\sqrt{t}$. The endpoint identifications in the join are exactly what is needed for this formula to descend to the quotient. We then prove that the descended map is continuous, bijective, and equivariant; compactness and Hausdorffness upgrade the continuous bijection to a homeomorphism. The index statement follows from equivariant invariance of the $\mathbb{Z}/2$-index together with the standard sphere-index theorem, namely that the antipodal sphere $S^n$ has $\mathbb{Z}/2$-index $n$. [/proofplan]

[step:Define the join quotient and the candidate sphere map] Let \begin{align*} Q = S^a \times S^b \times [0,1]. \end{align*} Define an [equivalence relation](/page/Equivalence%20Relation) $\sim$ on $Q$ by declaring \begin{align*} (x,y,0) \sim (x,y',0) \end{align*} for all $x \in S^a$ and all $y,y' \in S^b$, and \begin{align*} (x,y,1) \sim (x',y,1) \end{align*} for all $x,x' \in S^a$ and all $y \in S^b$, with no further identifications except those forced by reflexivity, symmetry, and transitivity. The join is the quotient [topological space](/page/Topological%20Space) \begin{align*} S^a * S^b = Q / \sim. \end{align*} Let \begin{align*} q: Q \to S^a * S^b \end{align*} be the quotient map, and write $[x,y,t] = q(x,y,t)$. Define \begin{align*} F: Q \to \mathbb{R}^{a+1} \times \mathbb{R}^{b+1} \end{align*} by \begin{align*} F(x,y,t) = \left(\sqrt{1-t}\,x,\sqrt{t}\,y\right). \end{align*} Since $|x|=1$ and $|y|=1$, we have \begin{align*} \left|\sqrt{1-t}\,x\right|^2 + \left|\sqrt{t}\,y\right|^2 = (1-t)|x|^2 + t|y|^2 = 1. \end{align*} Thus $F(Q) \subset S^{a+b+1}$, where $S^{a+b+1}$ is viewed as the unit sphere in $\mathbb{R}^{a+1} \times \mathbb{R}^{b+1}$. [/step]

[step:Descend the formula to a continuous map on the join]We prove that $F$ is constant on equivalence classes. If $t=0$, then \begin{align*} F(x,y,0) = (x,0) \end{align*} is independent of $y \in S^b$. Hence $F(x,y,0)=F(x,y',0)$ whenever $(x,y,0)\sim(x,y',0)$. If $t=1$, then \begin{align*} F(x,y,1) = (0,y) \end{align*} is independent of $x \in S^a$. Hence $F(x,y,1)=F(x',y,1)$ whenever $(x,y,1)\sim(x',y,1)$. Therefore there is a unique map \begin{align*} \Phi: S^a * S^b \to S^{a+b+1} \end{align*} satisfying \begin{align*} \Phi([x,y,t]) = \left(\sqrt{1-t}\,x,\sqrt{t}\,y\right) \end{align*} for every $(x,y,t)\in Q$. Since $F=\Phi\circ q$ and $F$ is continuous, the defining property of the [quotient topology](/page/Quotient%20Topology) implies that $\Phi$ is continuous.[/step]

[guided]The only possible obstruction to defining $\Phi([x,y,t])$ by the displayed formula is that a point of the join may have several representatives. We must check that the formula gives the same value for all representatives. The quotient relation only changes the $S^b$-coordinate when $t=0$ and only changes the $S^a$-coordinate when $t=1$. At $t=0$, the coefficient of $y$ is $\sqrt{0}=0$, so \begin{align*} \left(\sqrt{1-0}\,x,\sqrt{0}\,y\right) = (x,0). \end{align*} This is independent of $y$. Hence all representatives $(x,y,0)$ with the same $x$ have the same image. At $t=1$, the coefficient of $x$ is $\sqrt{0}=0$, so \begin{align*} \left(\sqrt{1-1}\,x,\sqrt{1}\,y\right) = (0,y). \end{align*} This is independent of $x$. Hence all representatives $(x,y,1)$ with the same $y$ have the same image. Thus the formula is well-defined on equivalence classes, giving a unique map \begin{align*} \Phi: S^a * S^b \to S^{a+b+1} \end{align*} with \begin{align*} \Phi([x,y,t]) = \left(\sqrt{1-t}\,x,\sqrt{t}\,y\right). \end{align*} Continuity follows from the quotient topology: the map \begin{align*} F: S^a \times S^b \times [0,1] \to S^{a+b+1} \end{align*} defined by the same formula is continuous, and it factors as $F=\Phi\circ q$, where $q$ is the quotient map. By definition of the quotient topology, this factor map $\Phi$ is continuous.[/guided]

[step:Recover the join coordinates from a point of the target sphere] We prove that $\Phi$ is bijective. Let $(u,v) \in S^{a+b+1} \subset \mathbb{R}^{a+1}\times\mathbb{R}^{b+1}$. Define \begin{align*} t = |v|^2. \end{align*} Since $|u|^2+|v|^2=1$, we also have $1-t=|u|^2$. If $0<t<1$, define \begin{align*} x = \frac{u}{|u|} \in S^a \end{align*} and \begin{align*} y = \frac{v}{|v|} \in S^b. \end{align*} Then \begin{align*} \Phi([x,y,t]) = \left(\sqrt{1-t}\,\frac{u}{|u|},\sqrt{t}\,\frac{v}{|v|}\right) = (u,v), \end{align*} because $\sqrt{1-t}=|u|$ and $\sqrt{t}=|v|$. If $t=0$, then $|v|=0$, so $v=0$ and $|u|=1$. For any fixed $y_0 \in S^b$, we have \begin{align*} \Phi([u,y_0,0]) = (u,0) = (u,v). \end{align*} If $t=1$, then $|u|=0$, so $u=0$ and $|v|=1$. For any fixed $x_0 \in S^a$, we have \begin{align*} \Phi([x_0,v,1]) = (0,v) = (u,v). \end{align*} Thus $\Phi$ is surjective. For injectivity, suppose \begin{align*} \Phi([x,y,t]) = \Phi([x',y',t']). \end{align*} Taking squared Euclidean norms in the second coordinate gives $t=t'$. If $0<t<1$, then both coordinate coefficients are positive, so \begin{align*} \sqrt{1-t}\,x = \sqrt{1-t}\,x' \end{align*} and \begin{align*} \sqrt{t}\,y = \sqrt{t}\,y', \end{align*} hence $x=x'$ and $y=y'$. Therefore $[x,y,t]=[x',y',t']$. If $t=0$, the equality of first coordinates gives $x=x'$, and the join relation identifies all choices of $y$ over the same $x$ at $t=0$. Hence $[x,y,0]=[x',y',0]$. If $t=1$, the equality of second coordinates gives $y=y'$, and the join relation identifies all choices of $x$ over the same $y$ at $t=1$. Hence $[x,y,1]=[x',y',1]$. Thus $\Phi$ is injective. [/step]

[step:Upgrade the bijection to a homeomorphism] The spaces $S^a$, $S^b$, and $[0,1]$ are compact Hausdorff spaces, so their product $Q=S^a\times S^b\times[0,1]$ is compact. Since $S^a*S^b$ is the quotient of $Q$, it is compact. The target $S^{a+b+1}$ is a Hausdorff subspace of the Euclidean space $\mathbb{R}^{a+1}\times\mathbb{R}^{b+1}$. A continuous bijection from a [compact space](/page/Compact%20Space) to a [Hausdorff space](/page/Hausdorff%20Space) is a homeomorphism. Applying this to $\Phi$, we conclude that \begin{align*} \Phi: S^a * S^b \to S^{a+b+1} \end{align*} is a homeomorphism. [/step]

[step:Verify equivariance under the antipodal actions] Let $\varepsilon$ denote the nonidentity element of $\mathbb{Z}/2$. For every $[x,y,t]\in S^a*S^b$, \begin{align*} \Phi(\varepsilon\cdot[x,y,t]) = \Phi([-x,-y,t]). \end{align*} Using the definition of $\Phi$, this equals \begin{align*} \left(\sqrt{1-t}\,(-x),\sqrt{t}\,(-y)\right) = -\left(\sqrt{1-t}\,x,\sqrt{t}\,y\right). \end{align*} The antipodal action on $S^{a+b+1}$ is multiplication by $-1$, so \begin{align*} \Phi(\varepsilon\cdot[x,y,t]) = \varepsilon\cdot\Phi([x,y,t]). \end{align*} Therefore $\Phi$ is $\mathbb{Z}/2$-equivariant. [/step]

[step:Conclude the index computation from equivariant invariance] We have constructed a $\mathbb{Z}/2$-equivariant homeomorphism \begin{align*} \Phi: S^a*S^b \to S^{a+b+1}. \end{align*} The $\mathbb{Z}/2$-index is invariant under $\mathbb{Z}/2$-equivariant homeomorphism, so \begin{align*} \operatorname{ind}_{\mathbb{Z}/2}(S^a*S^b) = \operatorname{ind}_{\mathbb{Z}/2}(S^{a+b+1}). \end{align*} By the standard sphere-index theorem for the antipodal action, which states that $\operatorname{ind}_{\mathbb{Z}/2}(S^n)=n$ for every $n\ge 0$ (citing a result not yet in the wiki: antipodal sphere has $\mathbb{Z}/2$-index equal to its dimension), we obtain \begin{align*} \operatorname{ind}_{\mathbb{Z}/2}(S^a*S^b) = a+b+1. \end{align*} This proves both the equivariant homeomorphism statement and the asserted index formula. [/step]