Is $D^4\cup_\phi D^3\times S^1$ a manifold

Question

Let $M:=D^4\cup_\phi D^3\times S^1$, where $\phi$ is the smashing product, i.e.$\phi:S^2\times S^1\overset{\wedge}\to S^3$.

Q:

• Is $M$ a manifold, since $\phi$ is not a smooth.

• Can we find a smooth map $f:S^2\times S^1\to S^3$ which is homotopic to $\phi$.

Answer 1

For your first question you should check whether the resulting space satisfies Poincaré duality; not all topological spaces do, but all topological manifolds do.

For your second question the answer is yes, as seen here. This is a standard result in differential topology.