I am studying a topology text, and got stuck on a problem.
Let $X$ be a union of three unit circles centered at $(-2, 0)$, $(0, 0)$ and $(2, 0)$. How do I show that $X$ is homotopy equivalent to a bouquet of three circles?
The book gives a hint to consider another problem that allows me to move three points, and I think it might be possible to say that both spaces are deformation contract of $\mathbb R^2 \setminus \{x, y, z \}$ for some $\{x, y, z \} \subset \mathbb R^2$, but not sure how rigorously argue it.
Can someone please correct me if I am wrong, and provide a detailed solution? The textbook lacks examples showing spaces are homotopy equivalent, and I am planning to do other problems that ask me to find homotopy equivalent.