Is $Y$ homeomorphic to $\mathbb S^1$? 
Let $Y = \mathbb S^1 \cup\{ (x,y): (x-2)^2 + y^2 =1 \}$ be a subspace of $\mathbb R^2$. 
  
  
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*Is $Y$ homeomorphic to $\mathbb S^1$?
  
*Is $Y$ homeomorphic to an interval?
  

Can anybody please help me with these? I am still very new to homeomorphisms.
 A: Your connectedness argument shows that $Y$ cannot be homeomorphic to $S^1$, but doesn't show that $Y$ cannot be homeomorphic to an interval. Let's assume such a homeomorphism $f:Y \to I$ exists. Note that $Y$ is compact and connected, so if it has any chance of being homeomorphic to an interval, the latter must be of the form $I = [a,b]$. Removing $a,b$ from $I$ yields a connected space, and for this to hold on the side of $Y$ we must remove one point from each circle (and those points cannot be $(1,0)$). But now, remove the point $(1,0)$ on $Y\setminus \{f^{-1}(a),f^{-1}(b)\}$: the resulting space has four connected components. On the other hand, removal of any point on $(a,b)$ yields only two connected components.

As you can see from these rather simple examples, the problem of distinguishing non-homeomorphic spaces rapidly becomes difficult. But with some basic tools from algebraic topology, we can solve these problems almost effortlessly! The first homology group $H_1(X)$ (an Abelian group) associated to a space $X$ is the first of many algebraic invariants associated to homotopy classes of topological spaces. In particular, if $H_1(X)$ is not isomorphic to $H_1(Y)$, then $X$ and $Y$ are not homeomorphic. This lets us reduce the topological problem to an algebraic one. Intuitively, $H_1$ measures the "$1$-dimensional holes" in your space.


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*The interval $I$ is contractible, that is, it can be continuously deformed to a point. Therefore $H_1(I) = 0$.

*The circle essentially embodies what a "$1$-dimensional hole" is. Therefore $H_1(S^1) \cong \mathbb{Z}$ is the free Abelian group on one generator: this generator corresponds to "one complete turn around the circle."

*Your space $Y$ is a wedge $S^1 \vee S^1$ of two circles, i.e., two circles glued together by a point. These represent two holes, and therefore $H_1(S^1 \vee S^1) \cong \mathbb{Z} \oplus \mathbb{Z}$.


All these Abelian groups are pairwise non-isomorphic. Therefore, this yields a more conceptual explanation of why your spaces aren't homeomorphic!
