Suppose that $p :X \to B$ is a fibration and $Y$ is a contractible space. Show that every map $f:Y \to B$ has a lift $g:Y \to X$.
Let $H:Y \times I \to Y$ be the homotopy $c_{y_0}\simeq \operatorname{id}_Y$ where $c_{y_0}$ is the constant map $c_{y_0}(y)=y_0$.
Consider $f : Y \to B$. Composing we get a homotopy $$f \circ H : Y\times I \to B.$$
Now in order to use the homotopy lifting property of $p$ we need a map $$Y \to X$$ but given the maps here I cannot figure out how I can construct a map with codomain $X$. Is the problem statement missing some assumptions?
Best Answer
You are almost there! You've already constructed a homotopy $G=f\circ H$ between $f$ and the constant map $c:Y\to B,\ y\mapsto f(y_0)$. (I assume that $G(y,0)=c$ and $G(y,1)=f$ for convenience) This constant map is easier to work with because it admits a lifting to $X$ that is easy to construct. More specifically, we choose $x_0\in X$ such that $p(x_0)=f(y_0)$ and define $\tilde c:Y\to X$ such that $\tilde c(y)=x_0$, then $p\circ\tilde c=c$. (Note that this lift needs not be unique. Any point in the fiber above $f(y_0)$ works.)
We now use the fact that $p:X\to B$ has the homotopy lifting property with respect to all topological spaces. In particular, given this homotopy $G$ between $c$ and $f$ and $\tilde c$ with $p\circ\tilde c=c$, there exists a homotopy $\tilde G$ such that $p\circ \tilde G= G$ and $\tilde G|_{Y\times\{0\}}=\tilde c$.
A lifting of $f$ can be defined by $\tilde f= \tilde G|_{Y\times \{1\}}$.
One can check that $p\circ \tilde f=(p\circ \tilde G)|_{Y\times \{1\}}=G|_{Y\times\{1\}}=f$.
There is a slight little wrinkle. We need to be careful when we choose $x_0\in X$ that satisfies $p(x_0)=f(y_0)$, where $c_{y_0}\simeq \text{id}_Y$. This is possible if and only if $f(Y)\cap p(X)\neq \varnothing$.
Clearly if such $x_0$ exists, then $f(Y)\cap p(X)\neq \varnothing$. Conversely, if $f(Y)\cap p(X)\neq\varnothing$, then there exists $y_0\in Y, \ x_0\in X$ s.t. $p(x_0)=f(y_0)$. Since $Y$ is contractible, there is a homotopy $\text{id}_Y\simeq c_{y_1}$, where $c_{y_1}:y\mapsto y_1$ for all $y\in Y$. We also know that $Y$ is path-connected (being contractible), so $c_{y_1}\simeq c_{y_0}$, where the homotopy is constructed by sliding the image along a path joining $y_1$ and $y_0$.
The reason that this is crucial is because there exists fibrations which are non-surjective, but usually it is enough to deal with surjective fibrations as suggested by the first answer here. Maybe the question is assuming surjectivity, in which case we can safely choose $x_0$.