Fundamental group of $S^1\vee S^1$ I want to show that $\pi_1(S^1\vee S^1) = \mathbb{Z}*\mathbb{Z}. $ I know that it follows from the Seifert-van Kampen theorem, but we haven't talked about that in class. 
We are given the following hint:

Define a group homomorphism $\mathbb{Z}*\mathbb{Z}\to\pi_1(S^1\vee S^1)$ via the two inclusions $S^1\to S^1\vee S^1,$ and show that this is an isomorphism. 

We've also never talked about free products, so I am very confused as to how to define this homomorphism.
 A: The left hand side is the free group on two letters, say $a$, $b$.  In terms of generators and relations, this free group is generated by two elements $a$ and $b$ subject to no nontrivial relations.  A typical element of this group looks like $aba^{-2}b^3 a^{-1} b$, i.e., a word in $a$, $b$, $a^{-1}$, and $b^{-1}$, and the group operation is given by concatenation.  
Free groups also enjoy a very nice universal property.  In your case, 
$$\operatorname{Hom}_{\mathbf{Grp}}(\mathbb{Z} * \mathbb{Z}, \pi_1(S^1 \vee S^1)) \cong \operatorname{Hom}_{\mathbf{Set}}(\{1,2\}, \pi_1(S^1 \vee S^1)),$$ that is, a group homomorphism out of $\mathbb{Z} * \mathbb{Z}$ corresponds to the choice of any two elements in the codomain, and vice-versa.  
The two inclusions $S^1 \hookrightarrow S^1 \vee S^1$ precisely give you two elements on $\pi_1(S^1 \vee S^1)$, which is exactly the data of a homomorphism $\mathbb{Z} * \mathbb{Z} \to \pi_1(S^1 \vee S^1)$.  In terms of the generators given above, this homomorphism is determined by sending $a$ to one of the two loops and $b$ to the other loop.  
A: Present the group $\Bbb Z * \Bbb Z = \langle a,b \rangle$.  It suffices to define our map on the generators $a,b$.
Let $L(S^1 \vee S^1)$ denote the group of loops over composition, and $\sim$ be the homotopy equivalence relation so that $\pi_1(X) = L(X)/\sim$.  Let $f$ be a map that sends $a$ to a loop around the first copy of $S^1$ send $b$ to a loop around the second copy of $S^1$.  Our homomorphism is $\pi_\sim \circ f$.
The trick is to show that this is an isomorphism.  How you do so depends on your tool set.  It's likely you have a theorem which looks something like this:

Theorem 59.1: Suppose $X = U \cup V$, where $U,V$ are open sets of $X$.  Suppose that $U \cap V$ is path connected, and that $x_0 \in U \cap V$.  Let $i$ and $j$ be the inclusion mappings of $U$ and $V$ respectively into $X$.  Then the images of the induced homomorphisms
  $$
i_*: \pi_1(U,x_0) \to \pi_1(X,x_0) \qquad 
j_*: \pi_1(V,x_0) \to \pi_2(X,x_0)
$$
  generate $\pi_1(X,x_0)$ (Munkres, "Topology" (Second Edition))

As it applies to our problem: the key is that every loop through $x_0$ (the intersection of our two circles) can be written as the composition of finitely many loops that "simple loops", i.e. those that begin and end at $x_0$ without passing through it again.  With that, we'll see that the homomorphism is surjective.
not sure about injectivity, though.
