I'm currently studying algebraic topology and I have the following question.
There are two similar topological spaces as below.
Below is a union of two Topologist's sine curves, and a square is added in the above.
I guess they are not homotopy equivalent, but I cannot find the reasoning.
Can anybody give some explanation? Thank you in advance.
(Added)
Let $T = \{(x,sin(\frac{1}{x})):x \in (0,1)\} \cup \{0\} \times [-1,1]$ be the Topologist's sine curve in $\mathbb{R}^2$.
Similarly, $T' = \{(x,sin(\frac{-1}{x})):x \in (-1,0)\} \cup \{0\} \times [-1,1]$
Below : $T \cup T'$
Above : $\{0\} \times [-1,1]$ is replaced by $[-1,1] \times [-1,1]$, and sine curve is translated accordingly.
They are not homotopy equivalent.
Let $$L = \{0\} \times [-1,1], \\ R = [-1,1]^2 , \\ S_\pm = \{(\pm x,\sin \frac 1 x) \mid x \in (0,1) \}, \\ S'_\pm = \{(\pm (x +1),\sin \frac 1 x) \mid x \in (0,1) \}.$$ We have to compare $$T = S_+ \cup S_- \cup L$$ and $$T' = S'_+ \cup S'_- \cup R .$$
Both spaces have three path components. Let $f : T \to T'$ be any map. It maps each of the path components $S_+, S_-,L$ of $T$ into one of the path components $S'_+,S'_-,R$ of $T'$.
Let $f^* : P \to P'$ the associated map between the set of path components. In order that $f$ be a homotopy equivalence, a necessary condition is that $f^*$ is a bijection. We shall show the assumption that $f^*$ is a bijection leads to a contradiction.
Case 1. $f(L) \subset S'_+$.
Since $f^*$ is bijective, we must have $f(S_+ \cup S_-) \subset S'_- \cup R$. The closure of $S_+ \cup S_-$ contains $L$, thus $f(L)$ must be contained in the closure of $f(S_+ \cup S_-)$ and thus contained in $S'_- \cup R$. This contradicts the assumption in case 1.
Case 2. $f(L) \subset S'_-$.
This is treated similarly.
Case 3. $f(L) \subset R$.
Then $f(S_+)$ must be contained in one of $S'_\pm$. W.l.o.g. we may assume $f(S_+) \subset S'_+$. It follows that $f(S_-) \subset S'_-$.
Since $L$ is contained in the closure of both $S_\pm$, we see that $f(L)$ must be contained in the closure of both $f(S_\pm)$ and thus in the closure of both $S'_\pm$. But these closures are disjoint, a contradiction.