Loch Ness monster and Jacob's Ladder Surfaces are NOT homeomorphic

I am looking for proof of the fact that $X($Jacob's Ladder surface$)$ is not homeomorphic to $Y($Loch Ness monster surface$)$. Certainly, fundamentals groups do not help as the fundamental groups of both spaces are free groups of countably many generators$($both $X,Y$ are homotopically equivalent to wedge of countably many circles$).$ Similarly, homology groups $H_0=\Bbb Z,H_1=\prod_{\text{countable}}\Bbb Z, H_2=0, H_n=0$ for all $n\geq 3$ in either cases.

Now, Ian Richard's classification of non-compact surfaces can be used to distinguish these spaces, looking at the number of ends. But I am looking for an alternative proof, if possible.

Any help will be appreciated. Thanks in advance.

Solution 1:

I will start with some very general remarks about proper homotopy invariants.

Let $X$ be a connected manifold (one needs much less). Consider an exhaustion of $X$ by compact submanifolds with boundary $K_i$: $$ X= \bigcup_i K_i, K_i\subset int(K_{i+1})~~ \forall i. $$ Every such exhaustion defines a direct system of maps of cohomology groups $$ H^*(X, X-K_i) \to H^*(X, X-K_j), i\le j. $$ The direct limit of this system is denoted $H_c^*(X)$, the compactly supported cohomology of $X$; it is independent of the exhaustion. Feeding the relative cohomology groups and their maps as above into long exact sequence of pairs $(X, X-K_i)$ we get a commutative diagram: $$ \begin{array}{ccccccccc} \to& \tilde{H}^{k-1}(X) & \to & \tilde{H}^{k-1}(X - K_i) & \to & H^k(X, X-K_i) & \to & \tilde{H}^{k+1}(X) & \to\\ & \downarrow & & \downarrow & &\downarrow& &\downarrow& \\ \to& \tilde{H}^{k-1}(X) & \to & \tilde{H}^{k-1}(X - K_j) & \to & H^k(X, X-K_j) & \to & \tilde{H}^{k+1}(X) & \to \end{array} $$ Taking the direct limit we obtain a long exact sequence $$ ... \to \tilde{H}^{k-1}(X) \to \tilde{H}^{k-1}_\epsilon(X) \to H_c^k(X) \to \tilde{H}^{k+1}(X) \to ... $$ The groups $\tilde{H}^{*}_\epsilon(X)$ are again independent of the exhaustion, they are direct limits of the systems $$ \tilde{H}^*(X-K_i) \to \tilde{H}^*(X-K_j), i\le j. $$ (You can see that they are independent of the exhaustion either by appealing to the independence of compactly supported cohomology groups or by repeating the same argument you use for $H^*_c$.)

Remark. This is actually quite general: If $(G_i)_{i\in I}$ is a direct system of groups (or, more generally, objects in some category) and $(G_i)_{i\in J}$ is a subsystem given by a cofinal subset $J\subset I$, then we get a natural isomorphism $$ \lim_{i\in J} G_i\cong \lim_{i\in I} G_i. $$ In our setting, $I$ will be the poset of all compact subsets of $X$ and $J\subset I$ will be a subset of $I$ given by a particular exhaustion $(K_i)$. The assumption that $(K_i)$ is an exhaustion implies that $J$ is cofinal in $I$.

Definition. The groups $\tilde{H}^*_\epsilon(X)$ are the reduced end-cohomology groups of $X$.

Remark. In fact, I did not need the compactly supported cohomology groups, I just wanted to relate the end-cohomology to something you already know.

Independence of the exhaustion implies that these groups are topological invariants of $X$; they are also invariants of proper homotopy type of $X$: Each proper homotopy-equivalence $X\to Y$ induces isomorphisms $$ H^*_c(Y)\to H^*_c(X), H^*_\epsilon(Y)\to H^*_\epsilon(X). $$

Now, back to your question. Take your surface $X$ and exhaust it by compact subsurfaces $K_i$ such that $X-K_i$ consists of two unbounded components. For $Y$, exhaust by compact subsurfaces $L_i$ each of which has connected (unbounded) complement. Computing the end-cohomology we get $$ \tilde{H}^0_\epsilon(X)= {\mathbb Z}, \tilde{H}^0_\epsilon(Y)=0 $$
since for each pair $j\ge i$ we get isomorphisms $$ {\mathbb Z}=\tilde{H}^0(X-K_i) \to \tilde{H}^0(X-K_j)= {\mathbb Z}, $$ $$ 0=\tilde{H}^0(Y-L_i) \to \tilde{H}^0(Y-L_j)= 0. $$ Hence, $X$ is not homeomorphic to $Y$. The same proof shows that these surfaces are not properly homotopy-equivalent.

Lastly, even though you did not ask about it, using Richards' classification of surfaces one can prove even more:

Theorem. Two surfaces are properly homotopy-equivalent if and only if they are homeomorphic.

Solution 2:

Non-Answer: Here is some figures that would be helpful.

• An easy observation but interesting for me: One can find a loop that after removing it the bounded component of both are homeomorphic to torus minus open disk: figures (a) and (d). So removing one loop trick does not work.
• Unfortunately removing 3 loops also does not work. one can do same process (b-c-d) to (d) and construct bounded components like (a).

Solution 3:

This is not my own answer. I have stolen this argument from Lee Mosher's comment. There is already a full and good solution provided by Moishe Kohan. In case you lost in the comments and still want an alternative solution, you may read below.

If $f:X→Y$ is a homeomorphism and $C\subseteq X$ is a subset, then $f$ restricts to a homeomorphism from $C$ to $f(C)$, from $X−C$ to $Y−f(C)$, and from $\overline{X−C}$ to $\overline{Y−f(C)}$. So if $X−C$ has two components each noncompact, then $Y−f(C)$ must have two components each noncompact. So if $X$ and $Y$ are homeomorphic, and $X$ has the property of possessing a loop with two complementary components each having noncompact closure, then $Y$ has that same property.

We can find an embedding $ψ:\Bbb S^1↪X$ such that $X−\text{im } ψ$ has two components each having non-compact closures, but if $Y$ were homeomorphic to $X$, then $Y$ would also have the same property, but, removing any loop form $Y$ we have exactly one component with non-compact closure.

But, why removing any loop form $Y$ we have exactly one component with non-compact closure? We can proceed as follows.

Let $S_{g,1}$ be an embedded compact submanifold of $Y$ having $g$-genus and one boundary component. Let $γ:\Bbb S^1↪Y$ be an embedded loop in $Y$ contained in the interior of some $S_{g,1}$. Now, $Y\backslash S_{g,1}⊆Y\backslash γ$, and $Y\backslash S_{g,1}$ is connected, so $Y\backslash S_{g,1}$ is contained in one component of $Y\backslash γ$. That is to say all other components of $Y\backslash γ$, if exist, must be inside in $S_{g,1}$ and thus bounded.