Non-nullhomotopic map from $S^1$ to $\mathbb{R}^2$ without two points, but which becomes nullhomotopic when adding a point back to $\mathbb{R}^2$. 
Prove that there is an embedding $f\colon S^1 \to \mathbb{R}^2 \setminus \{a_1, a_2 \}$, where $a_1, a_2 \in \mathbb{R}^2, a_1 \neq a_2$ which is not nullhomotopic, but which becomes nullhomotopic when regarded as a function from $S^1$ to $\mathbb{R}^2 \setminus \{a_i\}$, for every $i \in \{1,2 \}$.

I don't really know how to prove this. For me, this question seems counter-intuitive.
I tried to use the lemma that says that if a path-connected space $X$ can be written as a union of path-connected subsets $X = A \cup B$ with $A \cap B \neq \emptyset$, then every loop based at $x \in A \cap B$ is homotopic to a concatenation of loops contained completely in $A$ or in $B$, but I have not been able to finish.
 A: I don't know if you can do this with an embedding $f\colon S^1 \to \mathbb{R}^2\setminus\{a_1, a_2\}$, I think any such map cannot be injective. 
If $f$ doesn't need to be an embedding then I saw a cool talk about this once, you can actually inform your search for such a loop using the Seifert-van Kampen theorem! (There are a few details you should fill in.)
Suppose $a_1 = (-1,0)$ and $a_2 = (1,0)$. Then you can cover $X = \mathbb{R}^2 \setminus\{a_1, a_2\}$ with open sets $U_1 = ((-\infty, 1)\times \mathbb{R})\setminus \{a_1\}$ and $U_2 = ((-1, \infty)\times \mathbb{R}) \setminus\{a_2\}$ and the intersection $(-1, 1)\times \mathbb{R}$ is contractible and contains the basepoint $(0,0)$.
Seifert-van Kampen then says that $\pi_1(X)\cong \pi_1(U_1)*\pi_1(U_2)\cong \mathbb{Z}*\mathbb{Z}$. That is, if $\alpha_j$ is a generator of $\pi_1(U_j)$ then every element of $\pi_1(X)$ can be written as a word in $\alpha_1$, $\alpha_2$, and their inverses. We also have two inclusion maps $i_j\colon X \to \mathbb{R}^2 \setminus\{p_j\} \simeq U_j$ such that both compositions $\pi_1(U_j) \to \pi_1(X) \to \pi_1(\mathbb{R}\setminus\{p_j\})\cong \pi_1(U_j)$ are the identity. In fact, the homomorphisms $(i_j)_*\colon \pi_1(X) \to \pi_1(U_j)$ have an explicit description since $\pi_1(X)$ is a free product: it is defined by sending $\alpha_j$ to itself and $\alpha_i$ ($i\neq j$) to $1$.
Now, how does this apply to your problem? You want a loop $\gamma \neq 1 \in \pi_1(X)$ such that  both $(i_1)_*(\gamma) = (i_2)_*(\gamma) = 1$. Given our analysis of what these homomorphisms are doing, it follows that a word is in $ker (i_1)_*\cap ker (i_2)_*$ iff for both values of $j$ the elements $\alpha_j$ and $\alpha_j^{-1}$ appear the same number of times. All you need to do is come up with an example of such a word which represents a non-trivial loop in $\pi_1(X)$.
