Difference between $S^1 \vee S^1$ and $S^1 / \sim$ I'm currently self-studying algebraic topology and I have the feeling I have some misunderstandings about some very basic topics. As the title states I don't get the difference between two circles attached at one point and one circle with two points identified.
I know that $\pi_1(S^1 \vee S^1) \simeq \mathbb{Z} \star \mathbb{Z}$ (attained e.g. by Van-Kampen) and $\pi_1(S^1 / \sim) \simeq \pi_1(S^1) \simeq \mathbb{Z}$
So obviously they are different. But why? If we're looking at pictures, both could be seen as the number 8. I guess the point where the two circles are joined is the important part.
If we look at $S^1 / \sim$ and take the spaces $A_1, A_2$ we get by taking two arcs $A^\prime_1, A^\prime_2 \subsetneq S^1$ that are "big enough" and then identifying the north and south pole, so we have (e.g.) the number 8 with the "west" and "east" pole removed. Both of these sets are pathconnected and open. They both contain the middle point and they cover $S^1 / \sim$. Their intersection is pathconnected in $S^1 / \sim$ (although not in $S^1$.).
So by this we should be able to apply Van-Kampens theorem and arrive at $\pi_1(S^1 / \sim) \simeq \mathbb{Z} \star \mathbb{Z}$. Obviously this is not the case, so I guess the problem is that $A^\prime_1 \cap A^\prime_2$ is not pathconnected in $S^1$. But why is that relevant? Please keep in mind I am working on the basics of algebraic topology, so I don't know anything about homology and the likes.
 A: Your claim that $\pi_1(S^1/\sim) \approx \mathbb Z$ is wrong.
Take any two distinct points $x, y \in S^1$ and form the quotient $S = S^1/\{x,y\}$ in which the two-point set $\{x,y\}$ is collapsed to a single point. This is precisely your space $S^1/\sim$. It is easy to check that $S$ is Hausdorff.
We shall understand $S^1 = \{ z \in \mathbb C \mid \lvert z \rvert = 1\}$. Then $S^1 \vee S^1$ can be defined as the quotient $(S^1 \times \{1,2\})/\sim$, where $\sim$ identifies the two points $(1,1)$ and $(1,2)$.
Write $x = e^{is}$ and $y = e^{it}$ with $s, t \in[0,2\pi)$. We have $s \ne t$ and w.l.o.g. we may assume $s < t$. Define
$$\phi : S^1 \vee S^1 \to S, \phi(e^{ir},k) = \begin{cases} e^{i(s+(t-s)r/2\pi)} & k = 1\\ e^{i(t+(s+2\pi-t)r/2\pi)} & k = 2 \end{cases}$$
You can easily verify that $\phi$ is a well-defined continous bijection. Since $S^1 \vee S^1$ is compact and $S$ is Hausdorff, $\phi$ is a homeomorphism. Thus your intuition that both spaces can be seen as the figure $8$ is correct.
