Homotopy of spaces $S^2\times S^4\setminus \{*\}$ and $S^2\vee S^4$ How to prove that $S^2\times S^4\setminus \{*\}$ and $S^2\vee S^4$ are homotopic?
I'm having trouble visualizing that. 
Thanks in advance.
 A: Sorry, this is not totally rigourous but I think the idea is correct. Consider a point $(p,q) \in S^2 \times S^4$. We'll retract $S^2 \times S^4 \backslash \{(-p,-q)\}$ to $\{p\} \times S^4 \cup S^2 \times \{q\}$ which is isomorphic to $S^2 \vee S^4$ wedged at $(p,q)$.
Take any point $(x,y)$ in $S^2 \times S^4 \backslash \{ (-p,-q) \}$ and take the geodesic $\gamma_{(x,y)}$ which relies $(-p,-q)$ to $(x,y)$. Define $r(x,y)$ = the closest point of $(x,y)$ in $\gamma_{(x,y)} \cap (\{p\} \times S^4 \cup S^2 \times \{q\})$. Notice that $r_{|\{p\} \times S^4 \cup S^2 \times \{q\}} = \text{id}$. It is clear by construction that $ \text{im}(r) = \{p\} \times S^4 \cup S^2 \times \{q\}$. Thus $(x,y)$ in $S^2 \times S^4 \backslash \{ (-p,-q) \}$  retracts on $S^2 \vee S^4$.
A: There is a map $\partial(e^{n+m}) =\partial (e^m \times e^n)=e^m \times \partial e^n \sqcup_{\partial e^m \times \partial e^n} \partial e^m \times e^n \to S^m \sqcup_\text{pt} S^n=S^m \vee S^n$.
This map is induced from the two maps $e^m \times \partial e^n \to e^m \to e^m/\partial e^m=S^m \hookrightarrow S^m \vee S^n$ and $\partial e^m \times e^n \to e^n \to e^n/\partial e^n=S^n \to S^m \vee S^n$.
The mapping cone $C(\partial(e^{n+m}) \to S^m \vee S^n)=S^m \times S^n$.
Exercise:  For any map $X \to Y$ and $x \in C(X)\setminus X$, $C(X \to Y) \setminus \{x\} $ is homotopy equivalent to $Y$.
Therefore $S^m \times S^n$ minus a point in the image of the interior of $e^{n+m}$ is homotopy equivalent to $S^m \vee S^n$.
