Nature of the intersection in Van Kampen

I am a little confused about how to understand the path connected intersection requirement of Van Kampen's Theorem.

Most examples I have seen for $$S^1 \vee S^1$$ take the following decomposition:

However, why not take the point of contact to be $$U\cap V$$ and $$U,V$$ to be the circles. $$U\cap V$$ would still be path connected, since it is a single point.

Am I correct and it just so happens that I haven't come across a point intersection example, or is there a reason why point intersection are not taken in applying the Van Kampen Theorem?

• Seifert-Van Kampen requires $U$ and $V$ (and therefore $U\cap V$ as well) to be open. Commented Oct 5, 2022 at 7:27
• Missed this bit, thanks @SassatelliGiulio! Could you please add it as an answer, so I can mark this as "accepted" Commented Oct 5, 2022 at 7:34
• I'd rather somebody else do it. Commented Oct 5, 2022 at 7:35

In the standard vesion of the Seifert - van Kampen theorem both $$U,V$$ are required to be open.
Let $$(Z,*) = (X_1,x_1) \vee (X_2,x_2)$$ be the one-point union of the two based spaces $$(X_i,x_i)$$. If $$\{x_i\}$$ is a strong deformation retract of an open neighborhood $$U_i \subset X_i$$, then $$\pi_1(Z,*) \approx \pi_1(X_1,x_1) * \pi_1(X_2,x_2)$$.
This clearly applies to $$S^1 \vee S^1$$.
Proof. Let $$V_1 = X_1 \vee U_2$$ and $$V_2 = U_1 \vee X_2$$. These are open subsets of $$Z$$ which cover $$Z$$. The intersection $$V_1 \cap V_2$$ is contractible. Now apply Seifert - van Kampen, noting that $$(V_i,*) \simeq (X_i,x_i)$$.