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Let $f,g: M \to M$ be two continuously isotopic homeomorphisms, where $M$ is an orientable manifold, not necessarily compact. We pick an orientation on $M$, where by an orientation I mean a continuous (in the sense explained here) choice of generators for the local homology groups $H_n(M, M - \{x \} )$, where $n = dim \, M$.

I want to prove that if $f$ is orientation preserving in the sense mentioned above, then $g$ must also be orientation preserving. The analogous result for diffeomorphisms and smooth isotopies is easy to prove, since we can use the continuity of $sgn \, (d_xf_t)$. However, I'm having some difficulty in the topological case:

First, it's not clear to me how to prove (assuming it's true, as in the smooth case) that the fact that $f$ is orientation preserving at a point $x \in M$ implies that it is orientation preserving at every point. That is, I'd like to prove that if for some $x$ in $M$ and $f_*: H_n(M, M - \{ x\}) \to H_n(M, M - \{ f(x)\}) $ we have that $f_*([M]_x) = [M]_{f(x)}$, then the same holds for every point in $M$ (here $[M]_x$ is our selected generator for $H_n(M, M-\{ x\})$).

Second, I'd have to show that $f_*([M]_x) = [M]_{f(x)}$ for some $x$ implies $g_*([M]_x) = [M]_{g(x)}$. If it were the case that for every $t \in [0,1] $, $f_t(x) = g(x) = f(x)$ then it would be easy, since we'd have $f_* = f_{t*} = g_*: H_n(M, M - \{ x\}) \to H_n(M, M - {f(x)}) = H_n(M, M - \{ g(x)\})$ by using the maps induced in relative homology by the $f_t: (M, M - \{ x\}) \to (M, M-\{ f(x)\})$.

Since that doesn't happen when $f(x) \neq g(x)$, it's not as simple (at least for me!) as looking at the induced maps and using homotopy invariance like above.

Any hints or comments would be appreciated. Thanks!

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First, it's not clear to me how to prove (assuming it's true, as in the smooth case) that the fact that $f$ is orientation preserving at a point $x \in M$ implies that it is orientation preserving at every point. That is, I'd like to prove that if for some $x$ in $M$ and $f_*: H_n(M, M - \{ x\}) \to H_n(M, M - \{ f(x)\}) $ we have that $f_*([M]_x) = [M]_{f(x)}$, then the same holds for every point in $M$ (here $[M]_x$ is our selected generator for $H_n(M, M-\{ x\})$).

This is true as long as $M$ is connected. For any $x\in M$, there is some open set $U$ containing $x$ and an element $[M]_U\in H_n(M,M-U)$ such that for all $y\in U$, the image of $[M]_U$ in $H_n(M,M-\{y\})$ is the generator $[M]_y$ chosen by our orientation. Shrinking $U$, we may assume that there is also such a class $[M]_{f(U)}\in H_n(M,M-f(U))$. We may further shrink $U$ to assume $U$ is an open ball in some neighborhood of $x$ homeomorphic to $\mathbb{R}^n$ so that the map $H_n(M,M-U)\to H_n(M,M-\{y\})$ is an isomorphism for each $y\in U$ (and similarly for $f(U)$) and so the elements $[M]_U$ and $[M]_{f(U)}$ are unique.

We then have that for any $y\in U$, $f$ sends $[M]_y$ to $[M]_{f(y)}$ iff $f_*:H_n(M,M-U)\to H_n(M,M-f(U))$ sends $[M]_U$ to $[M]_{f(U)}$. The latter condition does not depend on the point $y\in U$. Thus if $f$ is orientation-preserving at $x$, it is orientation-preserving in a neighborhood of $x$, and similarly if it is orientation-reversing. So the sets of points where $f$ is orientation-preserving and where $f$ is orientation-reversing are an open partition of $M$. If $M$ is connected, one of these sets must be all of $M$.

Second, I'd have to show that $f_*([M]_x) = [M]_{f(x)}$ for some $x$ implies $g_*([M]_x) = [M]_{g(x)}$.

It suffices to show that $f_*([M]_x) = [M]_{f(x)}$ implies $(f_t)_*([M]_x) = [M]_{f_t(x)}$ for all sufficiently small $t$. Fixing a neighborhood $V\cong\mathbb{R}^n$ of $f(x)$, there is an open neighborhood $U$ of $x$ and an $\epsilon>0$ such that $f_t(U)\subseteq V$ for all $t\in[0,\epsilon]$. We may as well assume $\epsilon=1$, $U$ is our entire domain, and $V$ is our entire codomain. That is, we may as well assume we have a homotopy consisting of open embeddings $f_t:U\to\mathbb{R}^n$ for each $t\in [0,1]$, and wish to show $f_0$ is orientation-preserving at $x$ iff $f_1$ is orientation-preserving at $x$.

But now we can modify our maps $f_t$ so that we can look at just one point: let $f_t'(y)=f_t(y)-f_t(x)$. Note that $f_t'$ is orientation-preserving at $x$ iff $f_t$ is orientation-preserving at $x$, since it differs by composition with a translation, which is orientation-preserving on $\mathbb{R}^n$. We also have $f_t'(x)=0$ for all $t$. So, by considering the induced maps $H_n(U,U-\{x\})\to H_n(\mathbb{R}^n,\mathbb{R}^n-\{0\})$, we see that $f_0'$ is orientation-preserving at $x$ iff $f_1'$ is orientation-preserving at $x$. It follows that $f_0$ is orientation-preserving at $x$ iff $f_1$ is orientation-preserving at $x$, as desired.

[In this argument I am using the fact that $\mathbb{R}^n$ has only two different orientations, so that when we orient the codomain $\mathbb{R}^n$ in the same way as our given orientation on $V$, we have either the standard orientation or its opposite. This guarantees that translation is orientation-preserving on $\mathbb{R}^n$, for our chosen orientation. The fact that $\mathbb{R}^n$ only has two orientations is because it is connected, using the first part of the answer above.]

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  • $\begingroup$ Thanks, that was very detailed and helpful! It had actually occurred to me that the isotopy could be modified like that when $M = \mathbb{R}^n$, but I didn't realize I could use that locally in the general case... Now I know! Thanks again. $\endgroup$ – Mauro Feb 4 '18 at 6:21

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