# Homotopy of continuous map from a space with finite fundamental group

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).

• 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/… . – Justin Young Dec 30 '18 at 13:30
• 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. – Pepper Dec 31 '18 at 14:10

## 1 Answer

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.

• 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. – Pepper Dec 29 '18 at 3:53
• 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 $g=(f_Y,1)$. – s.harp Dec 29 '18 at 18:04
• @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. – Eric Wofsey Dec 29 '18 at 18:10
• Ah, you are right. – s.harp Dec 29 '18 at 18:12