Suppose $X$ is a topological space. We define $X_+$ to be the based space whose underlying space is $X \sqcup *$ and $*$ is the base point. Provide an explicit homeomorphism between $S^n _+ \land S^1 $ (smash product) and the space obtained by identifying the north and the south poles of $S^{n+1}$.
I only know the defintion of Smash product. I tried but could not make any progress whtasoever.
Thanks in advance for help.
Observe that $\mathbb{S}^n_+\land \mathbb{S}^1=\frac{(\mathbb{S}^n\sqcup *)\times(\mathbb{S}^1,1)}{(\mathbb{S}^n\sqcup *)\lor(\mathbb{S}^1,1)}=\frac{\mathbb{S}^n\times\mathbb{S}^1\bigsqcup (*\times \mathbb{S}^1)}{\mathbb{S}^n\times 1\cup (*\times\mathbb{S}^1)}=\frac{\mathbb{S}^n\times\mathbb{S}^1}{\mathbb{S}^n\times 1}$. So we just have to check that $\frac{\mathbb{S}^n\times\mathbb{S}^1}{\mathbb{S}^n\times 1}\cong \frac{\mathbb{S}^{n+1}}{\{N=S\}}$, where $N, S$ are respectively the North and South poles.
To see this, observe that $\frac{\mathbb{S}^n\times\mathbb{S}^1}{\mathbb{S}^n\times 1}$ is the result of a quotient of $\mathbb{S}^n\times I$, first via the map $(Id_{\mathbb{S}^n},\exp)$ (which identifies $(x,0)=(x,1), \forall x\in \mathbb{S}^n$) to get $\mathbb{S}^n\times\mathbb{S}^1$, and then collapses the subspace $\mathbb{S}^n\times 1$ of all the points $\{(x,0),(x,1)\}_{x\in \mathbb{S}^n}$ (after the first quotient, $\{(x,0),(x,1)\}$ is a single point for every $x\in \mathbb{S}^n$).
On the other hand, $\frac{\mathbb{S}^{n+1}}{\{N=S\}}$ can also be seen as a quotient of $\mathbb{S}^n\times I$ by first collapsing each of the subspaces $\mathbb{S}^n\times 0$, $\mathbb{S}^n\times 1$ to get $\mathbb{S}^{n+1}$ and then identifying these two points that would be the North and South poles.
The thing is that through this two quotients, the fibers are the same, i.e in both cases the subspaces collapsed to points are the same in $\mathbb{S}^n\times I$. By the uniqueness of quotient spaces (e.g Theorem 3.75 page 72 in John Lee's Introduction to Topological Manifolds, 2011), you get your homeomorphism.
If you take $n=1$ and make both quotients, you'll get a very good feeling of what I've done here.