# 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 $g:Y \to X$.

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?

• Choose $z_0 \in X$ such that $p(z) = f(y_0)$ and then use $y \mapsto z_0: Y \to X$. Commented Dec 18, 2023 at 22:26

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 $$x_0$$.