Seeing $\mathbb{S}^3$ as a pullback

Question

$\require{AMScd}$ Using the Hopf Fibration $$ \mathbb{S}^1 \hookrightarrow \mathbb{S}^3 \rightarrow \mathbb{S}^2 $$ and the fibration $$\mathbb{S}^1 \hookrightarrow \mathbb{S}^\infty \rightarrow \mathbb{CP}^\infty $$

We have the following diagram \begin{CD} \mathbb{S}^1 @>{=}>> \mathbb{S}^1\\ @VVV @VVV\\ \mathbb{S}^3 @>{i}>> \mathbb{S}^\infty\\ @VVV @VVV\\ \mathbb{S}^2 @>{j}>> \mathbb{CP}^\infty \end{CD} with $i,j$ inclusions.

My question is: is it true that we can see $ \mathbb{S}^3 $ as the pullback of the following diagram? And how could I prove it?

$$ \mathbb{S}^2 \rightarrow^j \mathbb{CP}^\infty \leftarrow^i \mathbb{S}^\infty $$

Answer 1

Yes. $S^\infty \to \mathbb{C}P^\infty$ is the universal $S^1$ bundle, which is the unit circle bundle of the tautlogical $\mathbb{C}$-bundle. The bundle $S^1 \to S^3 \to S^2\cong \mathbb{C}P^1$ is just the restriction to the $2$-skeleton (i.e. the classifying map is just the cannonical inclusion $S^2 \to \mathbb{C}P^\infty$), in general the pullback to the $2n$ skeleton gives you the bundle $S^1 \to S^{2n+1} \to \mathbb{C}P^n$

Answer 2

A more nonsense answer is that since the right hand map $S^ \infty \rightarrow \mathbb{C}P^\infty $ is a fibration, the pullback is the same as the homotopy pullback. Since $S^\infty$ is contractible, this is the statement that $S^3 \rightarrow S^2 \rightarrow \mathbb{C}P^\infty$ is a fibration sequence. This follows from the fact that $S^3 \rightarrow S^2$ and the homotopy fiber of $S^2 \rightarrow \mathbb{C}P^\infty$ are 2-connected covers of $S^2$ which are unique.