show/hide this revision's text 3 added 290 characters in body

These answers look at bit complicated so maybe there is something obviously wrong with the following argument:

Every embedded two-sphere $\Sigma \subset S^2 \times {\mathbb R}^2$ is displaceable: there is a one-parameter group (or family) of homeomorphisms $\varphi_t$ from $S^2 \times {\mathbb R}^2$ to itself such that $\varphi_T (\Sigma)$ is disjoint from $\Sigma$ for some (large) $T$. Indeed, just translate in the second variable far enough.

However, it is impossible to displace the zero section of $TS^2$ because its self-intersection number is $2$.

I read somewhere that to distinguish homeomorphism type of homotopic spaces one could look at the homotopy invariants of configuration spaces. I wonder :

Is the homotopy type of the (two-point) configuration space $C_2(S^2 \times {\mathbb R}^2)$ different from that of $C_2(TS^2)?$.

Edit. It turns out that the answer to the preceeding question is yes as is nicely explained here by Paolo Salvatore. This provides yet another way of proving that $S^2 \times {\mathbb R}^2$ and $TS^2$ are not homeomorphic.
show/hide this revision's text 2 deleted 9 characters in body; deleted 1 characters in body

These answers look at bit complicated so maybe there is something obviously wrong with the following argument:

If $\Sigma$ is an

Every embedded two-sphere in $\Sigma \subset S^2 \times {\mathbb R}^2$ , then it is displaceable: there is a one-parameter group (or family) of homeomorphisms $\varphi_t$ from $S^2 \times {\mathbb R}^2$ to itself such that $\varphi_T (\Sigma)$ is disjoint from $\Sigma$ for some (large) $T$. Indeed, just translate in the second variable far enough.

However, it is impossible to displace the zero section of $TS^2$ because its self-intersection number is $2$.

I read somewhere that to distinguish homeomorphism type of homotopic spaces , one could look at the homotopy invariants of configuration spaces. I wonder :

Is the homotopy type of the (two-point) configuration space $C_2(S^2 \times {\mathbb R}^2)$ different from that of $C_2(TS^2)?$.
show/hide this revision's text 1

These answers look at bit complicated so maybe there is something obviously wrong with the following argument:

If $\Sigma$ is an embedded two-sphere in $S^2 \times {\mathbb R}^2$, then it is displaceable: there is a one-parameter group (or family) of homeomorphisms $\varphi_t$ from $S^2 \times {\mathbb R}^2$ to itself such that $\varphi_T (\Sigma)$ is disjoint from $\Sigma$ for some (large) $T$. Indeed, just translate in the second variable far enough.

However, it is impossible to displace the zero section of $TS^2$ because its self-intersection number is $2$.

I read somewhere that to distinguish homeomorphism type of homotopic spaces, one could look at the homotopy invariants of configuration spaces. I wonder :

Is the homotopy type of the (two-point) configuration space $C_2(S^2 \times {\mathbb R}^2)$ different from that of $C_2(TS^2)?$.