I'm working on a practice problem that I really need help with.

Given two semi-locally simply connected, path connected, and locally path connected spaces $X$ and $Y$, we know there are universal covering maps $c_x: E\to X$ and $c_y: E'\to Y$.

Here is the problem:

Let $f: X\to Y$ be continuous. Prove that we can find some map $\tilde f: E\to E'$ with $c_y\circ \tilde f = f\circ c_x$.

I know that $E$ and $E'$ are simply connected spaces. Hence there is a homeomorphism $h:E\to E'$ such that $c_x =c_y\circ h$.

Also, in Munkres, since $c_y$ is a covering map, there is a lemma that guarantees the existence of unique lift of $f$ to a map $g:X\to E'$ if and only if $g_\ast(\pi_1(X))\subset (c_y)_\ast(\pi_1(E'))$. And similarly for $c_x$.

But I am not sure how to put this together to get the map $\tilde f$ described above.

Also, for uniqueness, would this have something to do with the fact that the fibers of the covering map have the discrete topology? I'm unsure about this, but it seems like a possibility.

  • $\begingroup$ There is a typo in your question. Since $f:X\to Y$, and $c_x:E\to X$, the composition $c_x\circ f$ is not defined. You probably mean $f\circ c_x$. $\endgroup$ – Alexis Leroux-Lapierre Jun 7 '17 at 16:07
  • $\begingroup$ @AlexisLeroux-Lapierre Yes, thanks for pointing that out. Would you please help me solve the problem? I need to know it for an exam coming up. $\endgroup$ – user453073 Jun 7 '17 at 16:11
  • $\begingroup$ Are you sure your question wants you to use universal coverings? This seems pretty weird to me. It reduces too much of the problem to a trivial case. Also, it is not true that two simply connected spaces are homeomorphic (take $\mathbb{R}$ and $S^2$ for example). $\endgroup$ – Alexis Leroux-Lapierre Jun 7 '17 at 20:47
  • $\begingroup$ @AlexisLeroux-Lapierre I am sure. I think it's supposed to be trivial, but I am pretty confused with the material. Could you indicate how the solution is simplified? I would appreciate it, since I am struggling with the material. $\endgroup$ – user453073 Jun 7 '17 at 20:53

Note that this solution was made without using universal covering of the spaces $X$ and $Y$.

We are going the following fundamental results :

Theorem 4.1 (W. S. Massey, A basic course in algebraic topology): Let $(E,p)$ be a covering space of $X$, $e_0\in E$ and $x_0=p(e_0)$. Then, the homomorphism $$\pi_1(p):\pi_1(E,e_0)\to \pi_1(X,x_0)$$ is injective.

Theorem 5.1 (W. S. Massey, A basic course in algebraic topology): Let $(E,p)$ be a covering space of $X$, $Y$ a connected and arcwise connected space, $y_0\in Y$, $e_0\in E$ and $x_0=p(e_0)$. Given a map $$\varphi:(Y,y_0)\to (X,x_0),$$ there exists a lifting $$\tilde{\varphi}:(Y,y_0)\to (E,e_0)$$ if and only if $$\mathrm{Im}\:(\pi_1(\varphi))\subset\mathrm{Im}\:(\pi_1(p)).$$ Furthermore, the map $\tilde{\varphi}$ is unique.

(I changed the notation a bit to what I am more used to. Also note that Massey assumes every space to be arcwise connected and locally arcwise connected.)

As for your problem (a), it is a direct application of theorem 5.1. Let $(E,p_X)$ and $(E',p_Y)$ be covering spaces of $X$ and $Y$ respectively. Fix the points $e_0\in E$ and $e_0'\in E'$, denoting $x_0=p_X(e_0)$ and $y_0=p_Y(e_0')$. Suppose there exists a base point preserving map $$f:(X,x_0)\to (Y,y_0).$$ This can be done by choosing $y_0$ to be $f(x_0)$ and then choosing $e_0'$ accordingly.

Consider the basepoint-preserving map $$g=f\circ p_X:(E,e_0)\to (Y,y_0).$$ Then, $g:E\to Y$ lifts into a map $\tilde{g}:E\to E'$ if and only if $$\mathrm{Im}\:(\pi_1(f\circ p_X))\subset\mathrm{Im}\:(\pi_1(p_Y)).$$ The map $\tilde{g}$ is the required map since $$\tilde{g}\circ p_Y=g=p_X\circ f$$ by construction. If you draw the commutative diagrams related to these maps, it will become clear what we have done. The trick was to consider $E$ as our "$Y$" from the theorem.

As for question (b), once you have removed the topological context, then it becomes some diagram chasing. Applying the $\pi_1$ functor to your diagram of covering maps, you then have a commutative diagram of abelian groups. We will do this in a pure algebraic way. Let $$A=\pi_1(E,e_0),\quad B=\pi_1(E',e_0'),\quad C=\pi_1(X,x_0),\quad D=\pi_1(Y,y_0)$$ and $$f=\pi_1(\tilde{g}),\quad g=\pi_1(p_Y),\quad h=\pi_1(p_X),\quad k=\pi_1(f).$$ Sorry for the notation, it might be a bit confusing. Let's now tackle the problem algebraicaly.

Suppose you have $A,B,C,D$, four abelian groups and the following maps $$ f:A\to B,\qquad g:B\to D,\qquad h:A\to C,\qquad k:C\to D $$ with $g$ and $h$ being injective. We will show that $f$ injective is equivalent to $k$ injective. Suppose $k$ is injective. Then, pick an $a\in \ker f$ and since the diagram commutes, $g(f(a))=k(h(a))$. This is equivalent to $k(h(a))=0$, and it follows that $a=0$ by injectivity of $k$ and $h$. This shows that $f$ is injective since $\ker f=\lbrace 0\rbrace$. Unfortunatly, I am unable to show that $f$ injective implies $k$ injective. I have not used injectivity hypothese on the map $f:X\to Y$ that you are supposing. Maybe we should look in this direction.

I hope this helped.

  • $\begingroup$ Thank you. I am not quite following your reasoning for part (b). I am completely unfamiliar with category theory (but I don't think you're really relying on it too much). I am mainly confused with how you are removing the topological context and which groups exactly $A, B, C, D$ represent in my original problem. Could you please elaborate? $\endgroup$ – user453073 Jun 7 '17 at 20:01
  • $\begingroup$ It may clarify things if you use the names of the groups in my original problem instead of $A, B, C$, and $D$. $\endgroup$ – user453073 Jun 7 '17 at 20:04
  • $\begingroup$ I will edit my post. Thank you for your suggestion, I should have done this. $\endgroup$ – Alexis Leroux-Lapierre Jun 7 '17 at 20:04
  • $\begingroup$ Also, for part (a), did you prove that $\tilde g$ will actually exist in our case? $\endgroup$ – user453073 Jun 7 '17 at 20:30
  • $\begingroup$ If lifts if and only if the conditions on the induced maps noted above are verified. $\endgroup$ – Alexis Leroux-Lapierre Jun 7 '17 at 20:32

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