# Can a factor map be a Serre fibration?

Let $D_n$ be an $n$-disc. Is the factor map $p: D_n\to D_n/S^{n-1}\simeq S^n$ a Serre fibration, in other words, can any homotopy $F: [0,1]\times X\to S^n$ be lifted to $\tilde{F}: [0,1]\times X\to D^n$, if a lift $\tilde{f}_0$ of $F(0,\cdot)$ is given?

More generally, if $p(S^{n-1})=\{s_0\}$, can a homotopy $F: [0,1]\times (X,A)\to (S^n, s_0)$ be lifted to $[0,1]\times (X,A)\to (D_n, S^{n-1})$, if $\tilde{f}_0$ is given? Suppose that $X$ is nice, such as a CW or simplicial complex.

• Serre fibrations have homotopy equivalent fibers. The map $p$ has two classes of fibers, the fiber over $[S^n]$ is a sphere, but the fiber over any other point is just a single point. – Dan Rust May 15 '14 at 20:35
• Thanks a lot, Daniel. So, for some homotopy $F$, the $\tilde{F}$ wouldn't exist? Could you give me some reference please? – Peter Franek May 15 '14 at 20:41
• The last sentence of the section on the formal definition of a fibration on wikipedia mentions the homotopy equivalence. I think it may be the case that actually Serre fibrations (which are a larger class of maps than Hurewicz fibrations) only have 'weakly equivalent' fibers but that still leads to a contradiction in this case. – Dan Rust May 15 '14 at 20:46
• Thanks a lot, Daniel, I've found it exactly as you say in Spanier.. However, concerning my second paragraph (the pair version), couldn't that be true? I need it :-)) – Peter Franek May 16 '14 at 14:21