Covering map for two circles Given a bouquet of two circles $\mathbb S^1 \vee \mathbb S^1$, I want to find a covering map $q: E \to \mathbb S^1 \vee \mathbb S^1$, whose induced homomorphism $q_* : \pi_1(E, e) \to \pi_1(\mathbb S^1 \vee \mathbb S^1, q(e))$ satisfy below given $n \geq 2$:
(1) the image $q_* (\pi_1(E, e))$ inside $\pi_1(\mathbb S^1 \vee \mathbb S^1, q(e))$ is a free group with $n$ generators
(2) the image $q_* (\pi_1(E, e))$ is a normal subgroup of $\pi_1(\mathbb S^1 \vee \mathbb S^1, q(e))$ such that $[\pi_1(\mathbb S^1 \vee \mathbb S^1, q(e)) : q_* (\pi_1(E, e))] = n - 1$ ($|\pi_1(\mathbb S^1 \vee \mathbb S^1, q(e)) / q_* (\pi_1(E, e))|$)
If $n = 2$, I think I can simply take $E$ to be a bouquet of two circles and just let $q$ to be a homoemorphism. For $n > 2$, should I consider $q$ from a bouquet of $2 n$ circles? Could someone give me a walk through of this question please?
 A: Your idea of taking a bouquet on $2n$ circles unfortunately doesn't work; any covering space $X$ of a bouquet $(\bigvee_{i=1}^kS^1,\star)$ will be a directed graph where every vertex has $k$ incoming edges and $k$ outgoing edges, corresponding to the $k$ incoming and $k$ outgoing edges of $\star$. Indeed, a vertex of $X$ with degree not equal to $2k$ would have no neighborhood homeomorphic to a neighborhood of $\star$, thus preventing $X$ from being a covering space. (To see this, consider the number of connected components of a small neighborhood after removing the vertex.) Here is an alternative approach; we use the fact that $q_*$ is an injection.

Label the circles of $S^1\vee S^1$ as $a$ and $b$, and let $F=\pi_1(S^1\vee S^1)=\langle a,b\rangle$. Consider the map $\phi:F\rightarrow C_{n-1}$, where $C_{n-1}=\langle c\rangle$ is cyclic of order $n-1$, induced by $a,b\mapsto c$. By the first isomorphism theorem, the kernel of this map will be a normal subgroup of $F$ of index $n-1$; furthermore, the elements of $\ker\phi$ will be precisely those of form $a^{m_1}b^{n_1}\dots a^{m_k}b^{n_k}$, where $\sum_{i=1}^km_i+n_i$ is divisible by $n-1$. This subgroup corresponds to the fundamental group of a graph of the following form on $n-1$ vertices. (This illustration is of the case $n=5$.)

Such a graph can be equipped with the obvious map to $S^1\vee S^1$ that takes every vertex to $\star$; this will be a covering map by the remark in my first sentence. Furthermore, such a graph has $2(n-1)$ edges and maximal trees of length $n-2$, so its fundamental group will be free on $2(n-1)-(n-2)=n$ generators, as desired.

Incidentally, this photo comes from Hatcher, a great resource to know of if you are learning algebraic topology.

Edit: Here are some answers to the questions you asked in the comments. For question 0; yes, the graph I've shown is a topological space, with topology induced by the subspace topology in $\mathbb{R}^2$, satisfying the desired properties of your problem. Alternatively, if you prefer the language of cell complexes, then a "graph" is simply a $1$-dimensional cell complex.
For question 1, do you know the definition of a cyclic group? There is a cyclic group of every finite cardinality. The answer to question 2 requires knowledge of cyclic groups, so if you're not familiar with them you should first read the wikipedia article I've linked. In any case, this is the fleshed out version of my argument. Every element of $F=\langle a,b\rangle$ is of the form $a^{m_1}b^{n_1}\dots a^{m_k}b^{n_k}$ for some $m_i,n_i\in\mathbb{Z}$. We have $$\phi(a^{m_1}b^{n_1}\dots a^{m_k}b^{n_k})=\phi(a)^{m_1}\phi(b)^{n_1}\dots\phi(a)^{m_k}\phi(a)^{n_k}=c^{m_1}c^{n_1}\dots c^{m_k}c^{n_k}=c^{\sum_{i=1}^km_i+n_i},$$ which will equal the identity in $C_{n-1}$ if and only if $\sum_{i=1}^km_i+n_i$ is divisible by $n-1$.
The answer to question 3 makes more sense after you know the answer to question 4, so I'll answer question 4 first. The main theorem is that the fundamental group of a graph is a free group, generated by the edges of the graph minus the edges of (any) maximal tree in the graph. This is quite a core theorem on fundamental groups, and is genuinely invaluable if you want to study covering spaces of graphs like $S^1\vee S^1$. For a reference on this theorem, see section IV.4. here, and for a worked through example see example 1.22 in the Hatcher book I linked to above.
Finally, for question 4, the subgroup $\ker\phi<\langle a,b\rangle$ doesn't exactly correspond to the graph I've shown, it corresponds to the fundamental group of that graph. Using the theorem I've mentioned in the answer to question 3, it is straightforward to verify this. Before you've learned the theorem though, if you just want to see this fact intuitively, try just picking a vertex of the graph as a basepoint and drawing out loops along the edges of the graph, writing down $a$, $b$, $a^{-1}$, or $b^{-1}$ depending on which edge you've traversed and in which direction... what kind of paths do you get?
