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I am dealing with the following problem:

Let $X$ and $Y$ be two connected non-compact surfaces, possibly with boundary. Let $f_0,f_1:(X,x_0)\to (Y,y_0)$ be two proper maps with $f_{0*}=f_{1*}:\pi_1(X,x_0)\to \pi_1(Y,y_0)$, then $f_0$ is properly homotopic to $f_1$.

Thoughts: Since $Y$ is a non-compact surface, its universal cover is a convex subset of $\Bbb R^2$, i.e., $Y=K(G,1)$ for some group $G$. If one goes through the proof the Hatcher's proposition 1B.9. on page 90, then the construction of the homotopy is dependent on $f_0$ and $f_1$, i.e., homotopy itself is proper if we are given $f_0$ and $f_1$ as proper maps. enter image description here

Any help will be appreciated. Thanks in advance.

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This is (a) false, and (b) obviously false.

I will explain (b) by showing that your question fails in the first non-trivial example. (You should therefore try to think of examples...)

Take $X = Y = S^1 \times \Bbb R$, with $x_0 = y_0 = (1, 0)$. Set $f_0$ to be the identity map, and $f_1(x,t) = (x,-t)$. Then these clearly induce the same map on the fundamental group --- they are the identity on the central circle, and $X$ deformation retracts to its central circle --- but they are also obviously not properly homotopic, as $\text{End}(X)$ is a two-point set, and proper maps induce maps on $\text{End}(X)$ which are equal if the original maps are properly homotopic. But $f_0$ is the identity on $\text{End}(X)$ and $f_1$ swaps the two ends.

If you want a proper map of a surfaces to be properly homotopic to the identity, you should (a) assume it is the identity on $\pi_1$, (b) assume it preserves ends, (c) assume it induces the identity on the fundamental group at infinity of each end. (For instance, this last condition distinguishes between the identity on $\Bbb R^2$ and the reflection map on $\Bbb R^2$.)

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  • $\begingroup$ Could you tell me any source of the portion "If you want a proper map of a surfaces to be properly homotopic to the identity....." $\endgroup$
    – Sumanta
    Commented Dec 23, 2020 at 12:41
  • $\begingroup$ I'd rather not. Try proving it first for closed surfaces with finitely many points removed. Then try doing the general case using what you learned. $\endgroup$
    – j.q
    Commented Dec 23, 2020 at 13:32
  • $\begingroup$ Got it. So, you are telling me to go through the following proof: Let $K$ and $L$ be finite dimensional CW-complexes. Let $f : K \to L$ be a proper map. Then $f$ is a proper homotopy equivalence if and only if: $(1)$ $\underline\pi_0(f) :\underline\pi_0(K) \to \underline\pi_0(L)$ is a homeomorphism. $(2)$ For each $n$, $\pi_n(f):\pi_n(K,*)\to \pi_n(L,f*)$ is an isomorphism. $(3)$ For each $n$ and each end $[a]$ of $K$, $\underline\pi_n(f) : \pi_n(K,\underline a) \to \underline\pi_n(L,\underline{fa})$ is an isomorphism. $\endgroup$
    – Sumanta
    Commented Dec 23, 2020 at 14:24

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