Let $f: S^d\to S^d$ be a (continuous) map of degree $1$, that is, homotopic to identity. Is it true that the mapping cylinder of $f$ is homeomorpic to a "normal" cylinder $S^d\times [0,1]$?
This is obvious if $f$ is a homeomorphism and still true in some other cases; however, I don't see how to prove it in general. Thanks for advice.
Map the circle to a wedge of three circles and then map each of the three circles to $S^1$ by maps of degrees respectively $+1$, $+1$, and $-1$. Then the resulting map $S^1\to S^1$ has degree $1$. If the mapping cylinder were homeomorphic to an ordinary cylinder, then a neighborhood of a typical point of $S^1\times \{1\}$ would look like a letter Y-shape times an interval, which cannot occur in a manifold.
