Q: On the proof of $S^m *S^n \cong S^{n+m+1}$ 
$S^m*S^n\cong S^{n+m+1}$

The solution of the above statement is already in this post. But I have a bit of trouble understanding it. In the solution, 'In the first piece, $\sim$ does nothing except pinching the copy of $S^m\times\{0\}$ to a point' and then he cone $S^m$ which is homeomorphic to $D^{m+1}$. In that statement, I thought $S^m\times S^n\times [0,1/2]/S^m\times S^n\times\{0\}\sim S^m\times\{\text{pt}\}\cong S^m\times CS^n\cong S^m\times D^{n+1}$. So I think the correct statement is that the first piece is homeomorphic to $S^m\times D^{n+1}$. Am I correct?
Edit: I found in Wikipedia that the statement can be more general: If $A,B$ are topological spaces then $A*B\cong CA\times B\cup_{A\times B}CB\times A$. I think the proof can proceed similarly: By definition, $A*B = A\times B\times I/A\times B\times\{0\}\sim A, A\times B\times\{1\}\sim B$. Divide into two parts, we get $A\times B\times[0,1/2]/A\times B\times \{0\}\sim B\cup_{A\times B}A\times B\times[1/2,1]/A\times B\times \{1\}\sim A$. The first part is homeomorphic to $CA\times B$ and the second part is homeomorphic to $A\times CB$. Now the conclusion follows.
