Let $(X_\gamma)_{\gamma \in \Gamma}$ be a collection of topological spaces, and let $x_\gamma \in X_\gamma$ be a fixed point for each $\gamma$. Fix some $\alpha \in \Gamma$, and suppose that for $\gamma \neq \alpha$ $x_\gamma$ is a deformation retract of $X_\gamma$, let $H_\gamma$ denote the associated deformation retraction.
I want to prove that $X_\alpha$ is a deformation retract of the wedge sum $\bigvee_{\gamma} X_\gamma$.

Consider the map $$H\colon I \times \bigvee_\gamma X_\gamma \to \bigvee_\gamma X_\gamma,$$ with $(t,[x,\gamma]) \mapsto [(H_\gamma(t,x),\gamma)]$ for $\gamma \neq \alpha$, and $(t,[x,\alpha]) \mapsto [(x,\alpha)]$.

The only thing to prove is that $H$ is continuous. So, let $U$ be open in the wedge sum and let $\pi$ denote the quotient map $\coprod X_\gamma \to \bigvee X_\gamma$, then $U=\pi(\coprod U_\gamma)$, where $U_\gamma \subset X_\gamma$ open and equality for almost all indices. Then $H^{-1}(U)$ is the union over $\gamma\in \Gamma$ of $$A_\gamma:=\{(t,[x,\gamma] \text{ } \vert \text{ } [H_\gamma(t,x),\gamma] \in U]\}.$$ Now, how do I see that all of these are open? This doesn't even seem obvious if I assume $x_\gamma \notin U_\gamma$ for all $\gamma$ and $\Gamma$ finite, because then $A_\gamma=(id\times pr_\gamma) (H_\gamma^{-1}(U_\gamma))$, where $pr_\gamma\colon X_\gamma \to \bigvee X_\beta$ is the obvious (in general not open) map.

Help is highly appreciated.

  • 2
    $\begingroup$ The wedge product is confusingly denoted \vee. The smash product (which is very different) is written \wedge. $\endgroup$ – user98602 Jun 18 '16 at 17:58
  • $\begingroup$ @MikeMiller Thanks, I will fix $\endgroup$ – user348647 Jun 18 '16 at 17:59

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