# lifting property of universal cover of $\mathbb{RP}^n$

I have a question. If we have a map $f:\mathbb RP^n\rightarrow \mathbb RP^n$, then can we always lift it to a map $g:S^n\rightarrow S^n$ such that the diagram commutes? $$\begin{array} S^n & \stackrel{g}{\longrightarrow} & S^n\\ \downarrow{p} & & \downarrow{p} \\ \mathbb RP^n & \stackrel{f}{\longrightarrow} & \mathbb RP^n \end{array}$$ Here $S^n$ is the universal cover of $\mathbb RP^n$. I have another question. Does any map $h: Y\rightarrow X$ can always be lifted to $\hat h: Y\rightarrow\hat X$ where $\hat X$ is the universal cover of X?

Any help will be appreciated!

In general, the lifting property says that given a map $h:Y\to X$, and a covering map $p: X' \to X$, the map $h$ lifts to a map $\hat h: Y\to X'$ (with designated base point $y_0$ mapping to designated base point $x_0'$ above $x_0=h(y_0)$) if and only if the containment $f(\pi_1(Y,y_0))\subseteq p(\pi_1(X',x_0'))$ is satisfied. In particular, if $Y$ is simply connected then this will always be true, so the answer to your first question is yes. On the other hand, the answer to the second question is no: a simple counterexample is given by $Y=X=S^1$ with $h$ being the identity map; if $h$ lifted to a map $\hat h: S^1 \to \mathbb R$, then $\hat h$ would be null-homotopic, and hence $h=p \circ \hat h$ would be null-homotopic as well, a contradiction, since we know that in fact $h$ is not null-homotopic.
• I see. $\mathbb{R}p^n$ is simply connected, since $H_1$=$\mathbb Z_2$ so the fundamental group is trivial. So the lift always exists. Then f in the above diagram is the composition of p and the lifting map. Thank you very much! Mar 15 '15 at 5:42
• Actually $\mathbb {RP}^n$ isn't simply connected: its fundamental group is $Z_2$ (for $n>1$). But what matters here is that $S^n$ is simply connected, since that is the domain of the map that we are trying to lift. Mar 15 '15 at 6:08
• Oh yeah $Z_2$ is commutative so the commutator is trivial. I guess we want to construct a $\hat f$ such that $f=p\circ \hat f$. Since $S^n$ is simply connected, $p(\pi_1(X',x'))$ is always trivial. Then how we lift it? Thanks! Mar 15 '15 at 15:23
• In this case both $f(\pi_1(Y,y_0))$ and $p(\pi_1(X',x_0'))$ are trivial, since we also have $Y=S^n$ being simply connected, so the required containment is satisfied. If you are asking how the lift is actually constructed, it works like this: given a point $y\in Y$, choose a path from the base point $y_0$ to $y$; the image of this path under $h$ is a path from $x_0$ to $h(x)$, which lifts to a path from $x_0'$ to some point in $X'$ which we define as $\hat h(y)$. The containment condition on the fundamental groups can be used to ensure that $\hat h$ is a uniquely defined, continuous map. Mar 15 '15 at 15:43