= Solution
{c}
Let $W=(\mathbb N\times S^1)/A$, and choose in each <circle> the point opposite $o$:
$$
z_n=(0,2/n)\in C_n,\qquad B=\{z_n:n\geq1\}\subseteq X.
$$
In the <subspace topology> from the <Euclidean plane>, the <sequence> $z_n$ converges to $o$, which does not belong to $B$. Thus $B$ is not <closed> in $X$.
On the other hand, let $D=\bar f^{-1}(B)\subseteq W$. With the <quotient map> $q$ from part (i),
$$
q^{-1}(D)=\mathbb N\times\{(0,1)\}.
$$
This is <closed> in $\mathbb N\times S^1$: its <complement> is the <open set> $\mathbb N\times(S^1\setminus\{(0,1)\})$. The definition of the <quotient topology>, applied to <complements>, says that $D$ is <closed> in $W$. A <homeomorphism> would carry this <closed set> to the <closed set> $B$, a contradiction. Therefore \b[$\bar f$ is not a <homeomorphism>].
The obstruction occurs at the common point. The <open set> $W\setminus D$ contains the collapsed point, but its <image> $X\setminus B$ contains no <neighbourhood> of $o$, since $z_n\to o$. The <wedge sum> topology allows a separate small <neighbourhood> on each <circle>; a <neighbourhood> of $o$ in the <Hawaiian earring> must contain all sufficiently small whole <circles>.
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