Suppose that $p:X\rightarrow B$ is a Serre fibration. I want to prove that $p$ has the right lifting property with respect to all maps of the form:
$$S^{n-1}\times I\cup_{S^{n-1}\times\{0\}}D^n\times\{0\}\rightarrow D^n\times I$$
Such arrows can be obtained form the universal mapping property of an pushout. To prove the claim I think one has to use that $S^{n-1}\times I\cup_{S^{n-1}\times\{0\}}D^n\times\{0\}$ is homoemorphic to $D^n\times\{0\}$. Then one can use the usual property of a Serre fibration to get an lift. Is the idea good? If so, why is this true?
I think that from this also follows that a Serre fibration has the homotopy lifting property to every CW-complex $A$. Is this a corollary from this fact above? Does one have to use induction on the skeletons?
Thanks a lot.
You have already answered yourself for the first part, since the inclusion $i:D^n\times \{0 \} \to D^n\times I$ and the map $j:S^{n-1} \times I \cup_{S^{n-1}\times \{0 \}} D^n \times \{ 0\}$ are such that there exists an isomorphism $\alpha$ for which $j\circ \alpha = i$, hence you can find a lift for $j$ since by definition you have one for $i$.
For the second part it is perhaps cleaner and conceptually easier to study the problem in greater generality, by means of closure properties of classes of arrows in weak factorization systems. In particular, it follows that you have liftings for any relative CW-complex.