Problem source: 2c on the UMD January, 2018 topology qualifying exam, seen here https://www-math.umd.edu/images/pdfs/quals/Topology/Topology-January-2018.pdf I have an argument for it, but I am not at all sure that it is correct. In particular, I would like to know 1) if it is correct and 2) if there is a better way of doing it (especially if my way is incorrect).

Let X be a connected CW complex with $\pi_1(X)$ finite. Let Y be a CW complex. Let $f:X\to Y\times S^1$ be a continuous map. Show there exists a $g:X\to Y$ such that $f$ is homotopic to the map $(g,1):X\to Y\times S^1$.

I am fairly confident in the first part of the proof, ie that $|f_*\pi_1(X)|<\infty$ (although my group theory is rusty). Since $f$ is continuous, it induces a group homomorphism on $\pi_1(X)$, so 1) $f_*\pi_1(X)\simeq\pi_1(X)/Ker(f_*)$ and 2) $f_*\pi_1(X)\leq\pi_1(Y\times S^1)\simeq\pi_1(Y)\times \mathbb{Z}$. In particular, (1) gives that $f_*\pi_1(X)$ is finite, hence $\simeq H\times \{0\}$, where $H$ is some subgroup of $\pi_1(Y)$, by (2).

The part I am much less sure of (that it is either correct or necessary) is the remainder. Take $p:Y\times\mathbb{R}\to Y\times S^1$ by $p(y,x)=(y,e^{2\pi ix})$. This is a covering map, and $p_*\pi_1(Y\times\mathbb{R})=\pi_1(Y)\times\{0\}$, so $f$ lifts to some $\tilde{f}$. Since the image of any closed loop in $X$ is taken to a closed loop that does not wind around $S^1$ (else, we would get a generator for an infinite fundamental group), $\tilde{f}$ is homotopic to some $(g,0)$, which maps to $(g,1)$ under $p$.

I have not used the fact that X and Y are CW complexes anywhere, but the only theorem that I know offhand in the context of CW complexes, subsets, and homotopy of functions would be the Homotopy Extension Theorem, which wouldn't make sense here (since $f(X)$ is not necessarily a CW complex).

  • $\begingroup$ The map to $Y$ is a red herring, since maps into a product are determined by the maps into the factors. For a fancy proof using cohomology/homology see: math.stackexchange.com/questions/380383/… . $\endgroup$ – Justin Young Dec 30 '18 at 13:30
  • $\begingroup$ That's a very nice proof, although I'll need to make it to homology in my notes before I can properly parse what's going on. $\endgroup$ – Pepper Dec 31 '18 at 14:10

This is basically correct and is the natural argument to make but your argument at the end is unclear. The reason that $\tilde{f}$ is homotopic to some $(g,0)$ is simply that $\mathbb{R}$ is contractible, so you can take $\tilde{f}$ and contract its second coordinate to $0$ via a homotopy. Your explanation seemed to be mixing this up with the reason that the lift $\tilde{f}$ exists at all.

As for where the assumption that the spaces are CW-complexes comes in, you need something about $X$ to be able to say that the lift $\tilde{f}$ exists at all (the usual lifting theorems for covering spaces only work for sufficiently nice spaces). In particular, it suffices to know that $X$ is path-connected and locally path-connected, which follows if you know that $X$ is a connected CW-complex. The assumption that $Y$ is a CW-complex is completely unnecessary.

By the way, the fact that $f_*\pi_1(X)$ is finite is completely trivial and requires no group theory: it is the image of the finite set $\pi_1(X)$ under a function.

  • $\begingroup$ So, in the lifting portion of the argument, it follows as soon as there is a lift to $Y\times\mathbb{R}$? The comments about not winding around $S^1$ were holdovers from my first attempt, and things got a little muddled along the way.Also, thank you for pointing out that I implicitly used X path connected and locally path connected - I had forgotten that that was a requirement for the theorem I used for the existence of a lift. $\endgroup$ – Pepper Dec 29 '18 at 3:53
  • $\begingroup$ The following seems a much simpler argument, is it incorrect? Let $p: Y\times S^1\to S^1$ be the projection, it follows from $\pi_1(X)$ finite that $(p\circ f)_*$ is constant and $p\circ f$ is null-homotop. If $h_t$ is the homotopy contracting the map, then $(f_Y, h_t)$ is a homotopy between $f=(f_Y,f_{S^1})$ and $g=(f_Y,1)$. $\endgroup$ – s.harp Dec 29 '18 at 18:04
  • $\begingroup$ @s.harp: That's essentially the same argument when you fill in the details. In particular, to prove $p\circ f$ is nullhomotopic you need to lift it to the universal cover. So you're doing the same thing, just separating out the second coordinate of $f$ before lifting to a covering space instead of after. $\endgroup$ – Eric Wofsey Dec 29 '18 at 18:10
  • $\begingroup$ Ah, you are right. $\endgroup$ – s.harp Dec 29 '18 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.