# Mapping cone not homotopy equivalent to quotient space

My question is about a "non-example" to theorem 1.6 in chapter VII in Bredon.

We have an inclusion $i: A \to X$, with $A = \{0\} \cup \{1/n | n = 1,2,...\}$, and $X = [0,1]$. Then $X/A$ is a one-point union of an infinite sequence of circles with radii going to zero. $C_i$, the mapping cone, contains homeomorphs of circles joined along edges.

Bredon claims that because the circles in $C_i$ do not tend to a point, any prospective homotopy equivalence $X/A \to C_i$ would be discontinuous at the image of $\{0\}$ in $X/A$.

I don't find this last argument very compelling as it stands (although I "intuitively" see it's correct). Could anyone expand on it? Or perhaps supply a more algebraic proof showing that homology or fundamental groups are different?

Let's consider the fundamental group $\pi_1(C_i)$. Let $a_0$ be the result of collapsing $A$ in the mapping cone $C_i$. Let $f : I \to C_i$ be a loop at $a_0$. Since $I$ is compact, $f$ is uniformly continuous. Put $\epsilon < 1/2$ and choose the corresponding $\delta > 0$. Cover $I$ with a finite collection of intervals $\{I_j\}$ of length less than $\delta$. The image of each such interval $f(I_j)$ is confined to an open ball of radius $\epsilon < 1/2$. Thus, it cannot be a nontrivial loop (any nontrivial loop has to travel through $a_0$ and $X$, which involves a straight distance larger than $\epsilon$). Since $\{I_j\}$ is finite, it follows that $f$ must consist of finitely many loops. Hence $\pi_1(C_i)$ is countable.
$X/A$, on the other hand, is the Hawaiian earring. Its fundamental group is uncountable. To see why, let $X/A = \bigcup_{n=1}^\infty B_n$, a union of circles of decreasing radius. We have retractions $r_n : X/A \to B_n$ resulting from collapsing all but one circle to the origin. Each $r_n$ induces a surjection $\rho_n : \pi_1(X/A) \to \mathbb Z$. The product $\rho$ of all $\rho_n$ gives a homomorphism from $\pi_1(X/A)$ to $\prod_{n=1}^\infty \mathbb{Z}$. This homomorphism is surjective since for every sequence $\{k_n\} \in \prod_{n=1}^\infty \mathbb{Z}$ there is a loop $g$ in $X/A$ that goes around the circle $B_n$ for $k_n$ times during the interval $[1 - 1/n, 1 - 1/(n+1)]$. $g$ is clearly continuous at points other than the origin. The fact the circles $B_n$ become closer and closer to the origin is crucial for showing that $g$ is continuous at the origin. Every neighborhood of the origin contains all but finitely many circles. This makes it possible to show that $g$ is continuous at the origin, and hence everywhere.