# Tate circles and suspension

I was trying to teach myself some motivic homotopy. I came across this notion of Tate circle. My question might be rather trivial.

1) I realize that ($\mathbb{P}^1$,$1$) is equivalent to the homotopy push out of $\mathbb{A}^1$ $\leftarrow$ $\mathbb{G}_m$ $\rightarrow$ $\mathbb{A}^1$. But how is this equivalent to the suspension $S^1 \wedge \mathbb{G}_m$?

2)Is there a good way to write smash product categorically? I tried doing this by taking the product diagram and taking the cofiber of the map $X_+ \sqcup Y_+ \rightarrow X \times Y$ which is essentially the definition of smash product. But this is complicated and downright ugly to look at.

Note that $S^1$ is the homotopy pushout of $*\leftarrow S^0\rightarrow *$ and that the derived smash product preserve homotopy pushout. So $S^1\wedge \mathbb{G}_m$ is the homotopy pushout of $*\wedge \mathbb{G}_m\leftarrow S^0\wedge \mathbb{G}_m\rightarrow *\wedge\mathbb{G}_m$. But $*\wedge \mathbb{G}_m\simeq *\simeq\mathbb{A}^1$ and $S^0\wedge\mathbb{G}_m\simeq\mathbb{G}_m$ (very easy exercise). So $S^1\wedge\mathbb{G}_m$ is the pushout of $\mathbb{A}^1\leftarrow\mathbb{G}_m\rightarrow\mathbb{A}^1$ so this is $(\mathbb{P}^1,1)$.
Categorically, smash product is the left adjoint to the mapping space $\operatorname{map}_*$. This is also the cofiber of $A\vee B\rightarrow A\times B$. Maybe there is more. But for $S^1\wedge X$, there is certainly more to say : this is called the suspension of $X$ and this is the homotopy pushout of $*\leftarrow X\rightarrow *$ as we just saw.