- Define what a covering map is.
- Define the
*degree*of a covering map, and show that it is well-defined. - Suppose that $X$ is a path-connected space such that $π_1(X)$ is finite. Then show that any continuous map $f: X → S^1$ is null-homotopic.

- A continuous map $p: \tilde{X} → X$ is a covering map if $X$ and $\tilde{X}$ are non-empty path-connected spaces, and, for every $x ∈ X$, there exists an open set $U_x$ containing $x$ such that $p^{-1}(U_x)$ is a disjoint union of open sets $V_j$ for $j ∈ J$ such that $\left.p\right|_{V_j}: V_j → U_x$ is a homeomorphism for all $j ∈ J$.
- If $x, x' ∈ X$, then $\left|p^{-1}(x)\right|=\left|p^{-1}(x')\right|$ since $\left|p^{-1}(x)\right|$ is locally constant (it is constant on each $U_x$) and $X$ is path-connected. The degree of $p$ is $\left|p^{-1}(x)\right|$ for any $x ∈ X$.
- Let $p: ℝ → S^1$ be the universal cover of $S^1$. Since the only finite subgroup of $π_1(S^1) ≅ ℤ$ is $\{0\}$, the map $f_*: π_1(X) → π_1(S^1)$ is zero. Hence, by the existence of lifts, there is a map $\tilde{f}: X → ℝ$ such that $f=p ∘ \tilde{f}$. In $ℝ$, the map $\tilde{f}$ is homotopic to a constant map by a straight-line homotopy. Hence $f$ is also null-homotopic.

- Show that $χ(X)=1$ if $X$ is a finite tree.
- If $X$ is a finite, connected graph, then we know that $π_1(X)$ is a free group $F$. Determine the number of generators of $F$ in terms of $χ(X)$.
- If $p: \tilde{X} → X$ is a finite-degree covering map and $X$ is a finite graph, then show that this induces a finite graph structure on $\tilde{X}$, and compute $χ(\tilde{X})$.

- We prove this by induction on $v(X)$. It is true when $v(X)=1$, since then $e(X)=0$. For the inductive step, note that every finite tree has a vertex of valency $1$. If we remove this vertex and the incident edge, we decrease both $v(X)$ and $e(X)$ by one, so $v(X)-e(X)$ remains unchanged.
- From lectures, we know that $F$ is generated by the edges that do not lie in a maximal tree. By (i), a maximal tree has $v(X)-1$ edges, hence $F$ is generated by $e(X)-v(X)+1=1-χ(X)$ elements.
- The space $X$ can be given the structure of a graph by taking the vertices of $\tilde{X}$ to be the preimages of the vertices $V$ of $X$ under $p$. An edge of $X$ corresponds to a map $f: I → X$. We lift this from the points of $p^{-1}(V)$ to obtain the corresponding edges of $\tilde{X}$. If $d$ is the degree of $p$, then $v(\tilde{X})=d v(X)$ and $e(\tilde{X})=d e(X)$, hence $χ(\tilde{X})=d χ(X)$.

Consider the graph $X$ with vertex set $ℕ$, an edge between $n$ and $n+1$ and a loop edge at $n$ for every $n ∈ ℕ$.
We take $b:=0$. Let $f(n):=n+1$ on the vertices and extend to the edges by applying $f$ to the endpoints. The map $g$ acts on the vertices by $g(n+1):=n$ for $n ∈ ℕ$ and $g(0)=0$, and acts on edges by applying $g$ to the endpoints (so the edge $(0,1)$ goes to the loop-edge at 0$)$. Then $g f=\text{id}_X$ by construction. We obtain a maximal tree in $X$ by taking all the edges $(n, n+1)$ for $n ∈ ℕ$, so $π_1(X, b)$ is generated by loops obtained from going from 0 to $n$, then around the loop edge at $n$, then back from $n$ to 0 (Sheet 4, Question 5). We obtain generators of $π_1(X, f(b))$ analogously. Hence, $f_*$ is not surjective, as the element of $π_1(X, f(b))$ corresponding to the loop-edge at 0 is not in the image. The map $g_*$ is not injective, since the elements of $π_1(X, f(b))$ corresponding to the loop-edges at 0 and 1 map to the same element of $π_1(X, b)$. [Alternately, note that $g$ maps the edge $(0,1)$ to constant path $\{0\}$.]