Reduced suspension and unreduced suspension

Question

In May's "A concise course in Algebraic Topology" Chap 14 section 1, the author says $\Sigma (X_+)$ is $\Sigma X\vee S^1$ where $X$ is an unbased space and $X_+$ is the union of a disjoint basepoint and $X$ and $\vee$ is the wedge sum. Obviously, $\Sigma(X_+)$ is the reduced suspension and $\Sigma X$ is the unreduced one. Can anybody show why this is so?

Answer 1

For the unreduced suspension for the unbased space $X$, we have \begin{align*} \Sigma X&=(X\times I)/(X\times \{0\}\cup X\times \{1\})\\ &= (X\times (0,1))\amalg \{*_0,*_1\} \end{align*} The trouble here is that we have two distinguished points $*_0$ and $*_1$ and we need to reduce it to have just one basepoint in a canonical way.

If $X$ is contractible, I guess this is what @Grumpy Parsnip meant by saying "if $X$ is nice enough" in his comment, we have $X\times (0,1)\simeq (X\times (0,1))\amalg (0,1)$. Hence, \begin{align*} \Sigma X\vee S^1&= (X\times (0,1))\amalg (0,1)\amalg \{1\sim *_0 \sim *_1\}\\&\simeq(X\times (0,1))\amalg \{*\}\\&= (X_+\times S^1)/\{*\}\times S^1 \cup X_+\times \{1\}\\ &= X_+\wedge S^1 = \Sigma X_+ \quad \text{ reduced suspension} \end{align*} The author seems to state this just to provide an example that $\Sigma (X_+)\nsim (\Sigma X)_+$ unlike a cone $CX$ where $C(X_+)\simeq(CX)_+$ so having this just for $X$ contractible seems to be enough to serve its goal.

I do realize that I'm abusing notations by ignoring topological structures by only considering a set structure level but I hope this would not be a big trouble. Let me know if anyone finds any flaw in my argument. Thanks!