# Summary
Up to this point, we have used covering spaces primarily as a tool for computing fundamental groups. Now we turn things around and use the fundamental group as a tool for studying covering spaces.

To do this in any reasonable way, we must restrict ourselves to the case where $B$ is locally path connected. Once we have done this, we may as well require $B$ to be path connected as well, since $B$ breaks up into the disjoint open sets $B_\alpha$ that are its path components, and the maps $p^{-1}(B_\alpha) \to B_\alpha$ obtained by restricting $p$ are covering maps, by Theorem 53.2. We may as well assume also that $E$ is path connected. For if $E_\alpha$ is a path component of $E$, then the map $E_\alpha \to B_\alpha$ obtained by restricting $p$ is also a covering map. Therefore, one can determine all coverings of the locally path-connected space $B$ merely by determining all path components of $B$ !

For this reason, we make the following:
$\textbf{Convention:}$
Throughout this chapter, when we state that $p: E \rightarrow B$ is a covering map will include the assumption that $E$ and $B$ are locally path connected and path connected, unless specifically stated otherwise.

With this convention, we now describe the connection between covering spaces of $B$ and the fundamental group of $B$.

If $p: E \to B$ is a covering map, with $p(e_0) = b_0$, then the induced homomorphism $p_*$ is injective, by Theorem 54.6, so that
\begin{equation}
H_0 = p_*(\pi_1(E, e_0))
\end{equation}is a subgroup of $\pi_1(B, b_0)$ isomorphic to $\pi_1(E, e_0)$. It turns out that the subgroup $H_0$ determines the covering $p$ completely, up to a suitable notion of equivalence of coverings. This we shall prove in §79. Furthermore, under a (fairly mild) additional "local niceness" condition on $B$, there exists, for each subgroup $H_0$ of $\pi_1(B, b_0)$, a covering $p: E \to B$ of $B$ whose corresponding subgroup is $H_0$. This we shall prove in §82.

Roughly speaking, these results show that one can determine all covering spaces of $B$ merely by examining the collection of all subgroups of $\pi_1(B, b_0)$. This is the classical procedure of algebraic topology; one "solves" a problem of topology by reducing it to a problem of algebra, hopefully one that is more tractable.

Throughout the chapter, we assume the general lifting correspondence theorem, Theorem 54.6.

$\textbf{Theorem 53.2}$
Let $p: E \rightarrow B$ be a covering map. If $B_0$ is a subspace of $B$, and if $E_0 = p^{-1}(B_0)$, then the restriction of $p$ is a covering map.

$\textbf{Theorem 54.6}$
Let $p: E \rightarrow B$ be a covering map, and let $p(e_0) = b_0$.
$(a)$The homomorphism $p_*:$ $\pi_1(E, e_0) \rightarrow \pi_1(B, b_0)$ is injective.
$(b)$Let $H = p_*(\pi_1(E, e_0) )$. The lifting correspondence induces an injective map:
\begin{equation}
\phi': \pi_1(B, b_0)/H \rightarrow p^{-1}(b_0)
\end{equation}
which is bijective if $E$ is path connected.
$(c)$If $f$ is a loop in $B$ based at $b_0$, then $[f] \in H$ if and only if $f$ lifts to a loop in $E$ based at $e_0$.

## Equivalence of Covering Spaces
In this section, we show that the subgroup $H_0$ of $\pi_1(B, b_0)$ determines the covering $p: E \to B$ completely, up to a suitable notion of equivalence of coverings.

$\textbf{Definition.}$
Let $p: E \to B$ and $p': E' \to B$ be covering maps. They are said to be $equivalent$ if there exists a homeomorphism $h: E \to E'$ such that $p = p' \circ h$. The homeomorphism $h$ is called an $\textit{equivalence of covering maps}$ or an $\textit{equivalence of covering spaces}$.


Given two covering maps $p: E \to B$ and $p': E' \to B$ whose corresponding subgroups $H_0$ and $H_0'$ are equal, we shall prove that there exists an equivalence $h: E \to E'$. For this purpose, we need to generalize the lifting lemmas of §54.

$\textbf{Lemma (The General Lifting Lemma):}$
Let $p: E \rightarrow B$ be a covering map; let $p(e_0) = b_0$. Let $f: Y \rightarrow B$ be a continuous map, with $f(y_0) = b_0$. Suppose $Y$ is path connected and locally path connected. The map $f$ can be lifted to a map $\tilde{f}: Y \rightarrow E$ such that $\tilde{f}(y_0) = e_0$ if and only if
\begin{equation}
f_*(\pi_1(Y, y_0)) \subseteq p_*(\pi_1(E, e_0)).
\end{equation}Furthermore, if such a lifting exists, it is unique.

$\textbf{Proof:}$
If the lifting $\tilde{f}$ exists, then $f_*(\pi_1(Y, y_0)) = p_*(\tilde{f}(\pi_1(Y, y_0))) \subseteq p_*(\pi_1(E, e_0))$. This proves the "only if" part of the theorem.

Now we prove that if $\tilde{f}$ exists, it is unique. Given $y_1 \in Y$, choose a path $\alpha$ in $Y$ from $y_0$ to $y_1$. Take the path $f \circ \alpha$ in $B$ and lift it to a path $\gamma$ in $E$ beginning at $e_0$. If a lifting $\tilde{f}$ of $f$ exists, then $\tilde{f}(y_1)$ must equal the end point $\gamma(1)$ of $\gamma$, for $\tilde{f} \circ \alpha$ is a lifting of $f \circ \alpha$ that begins at $e_0$, and path liftings are unique.

Finally, we prove the "if" part of the theorem. The uniqueness part of the proof gives us a clue how to proceed. Given $y_1 \in Y$, choose a path $\alpha$ in $Y$ from $y_0$ to $y_1$. Lift the path $f \circ \alpha$ to a path $\gamma$ in $E$ beginning at $e_0$ and define $\tilde{f}(y_1) = \gamma(1)$. See Figure 79.1. It is a certain amount of work to show that $\tilde{f}$ is well-defined, independent of the choice of $\alpha$. Once we prove that, continuity of $\tilde{f}$ is proved easily, as we now show.

To prove continuity of $\tilde{f}$ at the point $y_1$ of $Y$, we show that, given a neighborhood $N$ of $\tilde{f}(y_1)$, there is a neighborhood $W$ of $y_1$ such that $\tilde{f}(W) \subseteq N$. To begin, choose a path-connected neighborhood $U$ of $f(y_1)$ that is evenly covered by $p$. Break up $p^{-1}(U)$ into slices, and let $V_0$ be the slice that contains the point $\tilde{f}(y_1)$. Replacing $U$ by a smaller neighborhood of $f(y_1)$ if necessary, we can assume that $V_0 \subseteq N$. Let $p_0: V_0 \to U$ be obtained by restricting $p$; then $p_0$ is a homeomorphism. Because $f$ is continuous at $y_1$ and $Y$ is locally path connected, we can find a path-connected neighborhood $W$ of $y_1$ such that $f(W) \subseteq U$. We shall show that $\tilde{f}(W) \subseteq V_0$; then our result is proved.

Given $y \in W$, choose a path $\beta$ in $W$ from $y_1$ to $y$. Since $\tilde{f}$ is well defined, $\tilde{f}(y)$ can be obtained by taking the path $\alpha * \beta$ from $y_0$ to $y$, lifting the path $f \circ(\alpha * \beta)$ to a path in $E$ beginning at $e_0$, and letting $\tilde{f}(y)$ be the end point of this lifted path. Now $\gamma$ is a lifting of $\alpha$ that begins at $e_0$. Since the path $f \circ \beta$ lies in $U$, the path $\delta = p^{-1} \circ f \circ \beta$ is a lifting of it that begins at $\tilde{f}(y_1)$. Then $\gamma * \delta$ is a lifting of $f \circ(\alpha * \beta)$ that begins at $e_0$; it ends at the point $\delta(1)$ of $V_0$. Hence $\tilde{f}(W) \subseteq V_0$, as desired.


Finally, we show $\tilde{f}$ is well defined. Let $\alpha$ and $\beta$ be two paths in $Y$ from $y_0$ to $y_1$. We must show that if we lift $f\circ \alpha$ and $f \circ \beta$ to paths in $E$ beginning at $e_0$, then these lifted paths end at the same point of $E$.

First, we lift $f(\alpha)$ to a path $\gamma$ in $E$ beginning at $e_0$; then we lift $f \circ \bar{\beta}$ to a path $\delta$ in $E$ beginning at the end point $\gamma(1)$ of $\gamma$. Then $\gamma * \delta$ is a lifting of the loop $f \circ(\alpha * \bar{\beta})$. Now by hypothesis,
\begin{equation}
p_*(\pi_1(Y, y_0)) \subseteq p_*(\pi_1(E, e_0)).
\end{equation}Hence $[f \circ (\alpha * \bar{\beta})]$ belongs to the image of $p_*$ , and Theorem 54.6 now implies that its lift $\gamma * \delta$ is a $loop$ in $E$.

It follows that $\bar{f}$ is well defined. For $\bar{\delta}$ is a lifting of $f \circ \beta$ that begins at $e_0$, and $\gamma$ is a lifting of $f \circ \alpha$ that begins at $e_0$, and both liftings end at the same point of $E$. #

$\textbf{Theorem}$
Let $p: E \rightarrow B$ and $p': E' \rightarrow B$ be covering maps; let $p(e_0) = p'(e'_0) = b_0$. There is an equivalence $h: E \rightarrow E'$ such that $h(e_0) = e'_0$ if and only if the groups
\begin{equation}
H_0 = p_*(\pi_1(E, e_0)) \quad \text{and} \quad H'_0 = p'_*(\pi_1(E', e'_0))
\end{equation}are equal. If $h$ exists, it is unique.

$\textbf{Proof:}$
We prove the "only if" part of the theorem. Given $h$, the fact that $h$ is a homeomorphism implies that
\begin{equation}
h_*(\pi_1(E, e_0)) = \pi_1(E', e'_0).
\end{equation}Since $p' \circ h = p$,we have $H_0 = H'_0$

Now we prove the "if" part of the theorem; we assume that $H_0 = H'_0$ and show that $h$ exists. We shall apply the preceding lemma . Consider the maps

Because $p'$ is a covering map and $E$ is path connected and locally path connected, there exists a map $h : E \to E'$ with $h(e_0) = e'_0$ that is a lifting of $p$ (that is, such that $p' \circ h = p$). Reversing the roles of $E$ and $E'$ in this argument, we see there is a map $k: E' \to E$ with $p \circ k = p'$. Now consider the maps
The map $k \circ h : E \to E$ is a lifting of $p$ (since $p \circ k \circ h = p' \circ h = p$), with $p(e_0) = e_0$.
The identity map $\text{Id}_E$ of E is another such lifting. The uniqueness part of the preceding lemma implies that $k \circ h = \text{Id}_E$. A similar argument shows that $h \circ k$ equals the identity map of $E'$ #

We seem to have solved our equivalence problem. But there is a somewhat subtle point we have overlooked. We have obtained a necessary and sufficient condition for
there to exist an equivalence $h : E \to E'$ that carries the point $e_0$ to the point $e_0'$. However, we have not yet determined under what conditions there exists an equivalence in general.

It is possible that there may be no equivalence carrying $e_0$ to $e_0'$, but that there is an equivalence carrying $e_0$ to some other point of $p'^{-1}(b_0)$. Can we determine whether this is the case merely by examining the subgroups $H_0$ and $H_{0'}$? We consider this problem now.

If $H_1$ and $H_2$ are subgroups of a group $G$, you may recall from algebra that they are said to be $conjugate$ subgroups if $H_2 = a H_1 a^{-1}$ for some element $a$ of $G$.Said differently, they are conjugate if the isomorphism of $G$ with itself that maps $x$ to $a x a^{-1}$ carries the group $H_1$ onto the group $H_2$. It is easy to check that conjugacy is an equivalence relation on the collection of subgroups of $G$. The equivalence class of the subgroup $H$ is called the $\textit{conjugacy class}$ of $H$.

$\textbf{Lemma}$
Let $p: E \to B$ be a covering map. Let $e_0$ and $e_1$ be points of $p^{-1}(b_0)$, and let $H_i = p_*(\pi_1(E, e_i))$.

$\textbf{(a)}$ If $\gamma$ is a path in $E$ from $e_0$ to $e_1$, and $\alpha$ is the loop $p \circ \gamma$ in $B$, then the equation $[\alpha] * H_1 * [\alpha]^{-1} = H_0$ holds; hence $H_0$ and $H_1$ are conjugate.

$\textbf{(b)}$ Conversely, given $e_0$ and given a subgroup $H$ of $\pi_1(B, b_0)$ conjugate to $H_0$, there exists a point $e_1$ of $p^{-1}(b_0)$ such that $H_1 = H$.

$\textbf{Proof}$
(a) First, we show that $[\alpha] * H_1 * [\alpha]^{-1} \subseteq H_0$. Given an element $[h]$ of $H_1$, we have $[h] = p_*([\tilde{h}])$ for some loop $\tilde{h}$ in $E$ based at $e_1$. Let $\tilde{h} = (\gamma * \tilde{h}) *\bar{\gamma}$; it is a loop in $E$ based at $e_0$, and
\begin{equation}
p_*([\tilde{k}]) = [(\alpha * h) * \bar{\alpha}] = [\alpha] * [h] * [\bar{\alpha}],
\end{equation}so the latter element belongs to $p_*(\pi_1(E, e_0)) = H_0$, as desired.

Now we show that $[\alpha] \cdot H_1 \cdot [\alpha]^{-1} \supseteq H_0$. Note that $\bar{\gamma}$ is a path from $e_1$ to $e_0$ and $\bar{\alpha}$ equals the loop $p \circ \bar{\gamma}$. By the result just proved, we have
\begin{equation}
[\alpha] * H_0 * [\alpha]^{-1} \subseteq H_1
\end{equation}which implies our desired result.

(b) To prove the converse, let $e_0$ be given and let $H$ be conjugate to $H_0$. Then $H_0 = [\alpha] \cdot H \cdot [\alpha]^{-1}$ for some loop $\alpha$ in $B$ based at $b_0$. Let $\gamma$ be the lifting of $\alpha$ to a path
in $E$ beginning at $e_0$, and let $e_1 = \gamma(1)$. Then (a) implies that $[\alpha] * H_1 * [\alpha]^{-1} = H_0$
We conclude that $H = H_1$.

$\textbf{Theorem}$
Let $p: E \rightarrow B$ and $p': E' \rightarrow B$ be covering maps; let $p(e_0) = p'(e'_0) = b_0$.The covering maps $p$ and $p'$ are equivalent if and only if the subgroups
\begin{equation}
H_0 = p_*(\pi_1(E, e_0)) \quad \text{and} \quad H'_0 = p'_*(\pi_1(E', e'_0))
\end{equation}of $\pi_1(B, b_0)$ are conjugate

$\textbf{Proof}$
If $h : E \rightarrow E'$ is an equivalence, let $e'_1 = h(e_0)$, and let $H_1' = p'_*(\pi_1(E', e'_0))$ above Theorem implies that $H_0 = H_1'$, while the preceding lemma tells us that $H_1'$ is conjugate to $H_0'$.

Conversely, if the groups $H_0$ and $H_0'$ are conjugate, the preceding lemma implies there is a point $e_1'$ of $E'$ such that $H_1' = H_0$.Above Theorem then gives us an equivalence $h : E \rightarrow E'$ such that $h(e_0) = e_1'$.

$\textbf{EXAMPLE 1.}$
Consider covering spaces of the circle $B = S^1$. Because $\pi_1(B, b_0)$ is abelian, two subgroups of $\pi_1(B, b_0)$ are conjugate if and only if they are equal. Therefore, two coverings of $B$ are equivalent if and only if they correspond to the same subgroup of $\pi_1(B, b_0)$.

Now, $\pi_1(B, b_0)$ is isomorphic to the integers $\mathbb{Z}$. What are the subgroups of $\mathbb{Z}$? It is a standard theorem of modern algebra that, given a nontrivial subgroup of $\mathbb{Z}$, it must be the group $Z_n$ consisting of all multiples of $n$ for some $n \in \mathbb{Z}_+$.

We have studied one covering space of the circle, the covering $p: \mathbb{R} \rightarrow S^1$. It must correspond to the trivial subgroup of $\pi_1(S^1, b_0)$ because $\mathbb{R}$ is simply connected. We have also considered the covering $p: S^1 \rightarrow S^1$ defined by $p(z) = z^n$, where $z$ is a complex number. In this case, the map carries a generator of $\pi_1(S^1, b_0)$ into $n$ times itself. Therefore, the group $\pi_1(S^1, b_0)$ corresponds to the subgroup $Z_n$ of $\mathbb{Z}$ under the standard isomorphism of $\pi_1(S^1, b_0)$ with $\mathbb{Z}$.

We conclude from the preceding theorem that every path-connected covering space of $S^1$ is equivalent to one of these coverings.

# The Universal Covering Space

Suppose $p : E \rightarrow B$ is a covering map with $p(e_0) = b_0$. If $E$ is simply connected, then $E$ is called a $\textit{universal covering space}$ of $B$. Since $\pi_1(E, e_0)$ is trivial, this covering space corresponds to the trivial subgroup of $\pi_1(B, b_0)$ under the correspondence defined in the preceding section. Theorem 79.4 thus implies that any two universal covering spaces of $B$ are equivalent. For this reason, we often speak of "the" universal covering space of a given space $B$. Not every space has a universal covering space, as we shall see. For the moment, we shall simply assume that $B$ has a universal covering space and derive some consequences of this assumption.

$\textbf{Lemma}$
Let $B$ be path connected and locally path connected. Let $p: E \to B$ be a covering map in the former sense (so that $E$ is not required to be path connected). If $E_0$ is a path component of $E$, then the map $p_0: E_0 \to B$ obtained by restricting $p$ is a covering map.

$\textbf{Proof}:$
We first show $p_0$ is surjective. Since the space $E$ is locally homeomorphic to $B$, it is locally path connected. Therefore, $E_0$ is open in $E$. It follows that $p(E_0)$ is open in $B$. We show that $p(E_0)$ is also closed in $B$, so that $p(E_0) = B$.

Let $x$ be a point of $B$ belonging to the closure of $p(E_0)$. Let $U$ be a path-connected neighborhood of $x$ that is evenly covered by $p$. Since $U$ contains a point of $p(E_0)$, some slice $V_\alpha$ of $p^{-1}(U)$ must intersect $E_0$. Since $V_\alpha$ is homeomorphic to $U$, it is path connected; therefore, it must be contained in $E_0$. Then $p(V_\alpha) = U$ is contained in $p(E_0)$, so that in particular, $x \in p(E_0)$.

Now we show $p_0: E_0 \rightarrow B$ is a covering map. Given $x \in B$, choose a neighborhood $U$ of $x$ as before. If $V_{\alpha}$ is a slice of $p^{-1}(U)$, then $V_{\alpha}$ is path connected; if it intersects $E_0$, it lies in $E_0$. Therefore, $p_0^{-1}(U)$ equals the union of those slices $V_\alpha$ of $p^{-1}(U)$ that intersect $E_0$; each of these is open in $E_0$ and is mapped homeomorphically by $p_0$ onto $U$. Thus, $U$ is evenly covered by $p_0$.

$\textbf{Lemma}$
Let $p$, $q$, and $r$ be continuous maps with $p = r \circ q$, as in the following diagram:


$\textbf{(a)}$ If $p$ and $r$ are covering maps, so is $q$.

$\textbf{(b)}$ If $p$ and $q$ are covering maps, so is $r$.

$\textbf{Proof:}$
(a) Assume that $p$ and $r$ are covering maps. We show first that $q$ is surjective. Given $y \in Y$, choose a path $\tilde{\alpha}$ in $Y$ from $y_0$ to $y$. Then $\alpha = r \circ \tilde{\alpha}$ is a path in $Z$ beginning at $z_0$; let $\tilde{\tilde{\alpha}}$ be a lifting of $\alpha$ to a path in $X$ beginning at $x_0$. Then $q \circ \tilde{\tilde{\alpha}}$ is a lifting of $\alpha$ to $Y$ that begins at $y_0$. By uniqueness of path liftings, $\tilde{\alpha} = q \circ \tilde{\tilde{\alpha}}$. Then $q$ maps the end point of $\tilde{\tilde{\alpha}}$ to the end point $y$ of $\tilde{\alpha}$. Thus, $q$ is surjective.

Given $y \in Y$, we find a neighborhood of $y$ that is evenly covered by $q$. Let $z = r(y)$. Since $p$ and $r$ are covering maps, we can find a path-connected neighborhood $U$ of $z$ that is evenly covered by both $p$ and $r$. Let $V$ be the slice of $p^{-1}(U)$ that contains the point $y$; we show $V$ is evenly covered by $q$. Let $\{U_{\alpha}\}$ be the collection of slices of $p^{-1}(U)$. Now $q$ maps each set $U_{\alpha}$ into the set $r^{-1}(U)$; because $A$ is connected, it must be mapped by $q$ into a single one of the slices of $r^{-1}(U)$. Therefore, $q^{-1}(V)$ equals the union of those slices that are mapped by $q$ into $V$. It is easy to see that each such slice is mapped homeomorphically onto $V$ by $q$. For let $p_0, q_0, r_0$ be the maps obtained by restricting $p$, $q$, and $r$, respectively, as indicated in the following diagram:


Because $p_0$ and $r_0$ are homeomorphisms, so is $q_0 = r_0^{-1} \circ p_0$.

$(b)$ Assume that $p$ and $q$ are covering maps. Because $p = r \circ q$ and $p$ is surjective, $r$ is also surjective.

Given $z \in Z$, let $U$ be a path-connected neighborhood of $z$ that is evenly covered by $p$. We show that $U$ is also evenly covered by $r$. Let $\{V_\beta\}$ be the collection of path components of $r^{-1}(U)$; these sets are disjoint and open in $Y$. We show that for each $\beta$, the map $r$ carries $V_\beta$ homeomorphically onto $U

Let $\{U_\alpha\}$ be the collection of slices of $p^{-1}(U)$; they are disjoint, open, and path connected, so they are the path components of $p^{-1}(U)$. Now $q$ maps each $U_\alpha$ into the set $r^{-1}(U)$; because $U_\alpha$ is connected, it must be mapped by $q$ into one of the sets $V_\beta$. Therefore, $q^{-1}(V_\beta)$ equals the union of a subcollection of the collection $\{U_\alpha\}$.Previous Lemma implies that if $U_{\alpha_0}$ is any one of the path components of $q^{-1}(V_\beta)$ then the map $q_0 : U_{\alpha_0} \rightarrow V_\beta$ obtained by restricting $q$ is a covering map.

==In particular, $q_0$ is surjective. Hence $q_0$ is a homeomorphism==, being continuous, open, and injective as well. Consider the maps


obtained by restricting $p$, $q$, and $r$. Because $p_0$ and $q_0$ are homeomorphisms, so is $r_0$.

$\textbf{Theorem 80.3.}$
Let $p: E \rightarrow B$ be a covering map, with $E$ simply connected. Given any covering map $r: Y \rightarrow B$, there is a covering map $q: E \rightarrow Y$ such that $r \circ q = p$.


This theorem shows why $E$ is called a $universal$ covering space of $B$; it covers every other covering space of $B$.

$\textbf{Proof}$
Let $b_0 \in B$; choose $e_0$ and $y_0$ such that $p(e_0) = b_0$ and $r(y_0) = b_0$. We apply The General Lifting Lemma to construct $q$. The map $r$ is a covering map, and the condition
\begin{equation}
p_*(\pi_1(E, e_0)) \subseteq r_*(\pi_1(Y, y_0)).
\end{equation} is satisfied trivially because $E$ is simply connected. Therefore, there is a map $q : E \to Y$ such that $r \circ q = p$ and $q(e_0) = y_0$. It follows from the preceding lemma that $q$ is a covering map. #

Now we give an example of a space that has no universal covering space. We need the following lemma.

$\textbf{Lemma}$
Let $p : E \to B$ be a covering map; let $p(e_0) = b_0$. If $E$ is simply connected, then $b_0$ has a neighborhood $U$ such that the inclusion $i : U \to B$ induces the trivial homomorphism.
\begin{equation}
i_* : \pi_1(U, b_0) \to \pi_1(B, b_0)
\end{equation}

$\textbf{Proof.}$
Let $U$ be a neighborhood of $b_0$ that is evenly covered by $p$; break $p^{-1}(U)$ up into slices; let $U_\alpha$ be the slice containing $e_0$. Let $f$ be a loop in $U$ based at $b_0$. Because $p$ defines a homeomorphism of $U_\alpha$ with $U$, the loop $f$ lifts to a loop $\tilde{f}$ in $U_\alpha$ based at $e_0$. Since $E$ is simply connected, there is a path homotopy $\tilde{F}$ in $E$ between $\tilde{f}$ and a constant loop. Then $p \circ \tilde{F}$ is a path homotopy in $B$ between $f$ and a constant loop.

$\textbf{EXAMPLE 1.}$
Let $X$ be our familiar "infinite earring" in the plane. If $C_n$ is the circle of radius $1/n$ in the plane with center at the point $(1/n, 0)$, then $X$ is the union of the circles $C_n$. Let $b_0$ be the origin; we show that if $U$ is any neighborhood of $b_0$ in $X$, then the homomorphism of fundamental groups induced by inclusion $i: U \rightarrow X$ is not trivial.

Given $n$, there is a retraction $r : X \to C_n$, obtained by letting $r$ map each circle $C_i$, for $i \neq n$ to the point $b_0$. Choose $n$ large enough that $C_n$ lies in $U$. Then in the following
diagram of homomorphisms induced by inclusion, $j_*$ is injective, hence $i_*$ cannot be trivial.


It follows that even though $X$ is path connected and locally path connected, it has no
universal covering space.