# Induced maps in local homology and isotopies

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!

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.]

• 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. – Mauro Feb 4 '18 at 6:21