Given a set $X$ and a disjoint collection $R \subseteq \operatorname{Pot}(X)$, define a relation $\sim$ on $X$ via $x \sim y \Leftrightarrow x = y \text{ or } \exists S \in R: x \in S \text{ and } y \in S$. Define $X/R = X/\sim$ as the set of equivalence classes. Also, if $X$ is equipped with a topology, equip $X/R$ with the quotient topology.
Consider $S^1 \subseteq \mathbb{C}$.
For each $t \in [0, 1]$, let $S_t = S^1$ be a copy of the circle, one for each point of $[0, 1]$. Let $A = \bigsqcup_{t \in [0, 1]} S_t$ be their disjoint union. Let $B = A/\{\{1 \in S_t \mid t \in [0, 1]\}\} $. This means that $B$ is obtained by gluing all circles in $A$ at one point. Let $C = ([0, 1] \sqcup A)/\{\{t \in [0, 1], 1 \in S_t\} \mid t \in [0, 1]\}$. This means that $C$ is obtained by gluing a point of each circle in $A$ to a point of $[0, 1]$.
Is the mapping $\varphi: C \to B, t \in [0, 1] \mapsto 1 \in B, z \in S_t \setminus \{1\} \mapsto z \in S_t \setminus \{1\}$ that contracts $[0, 1]$ to the point a homotopy equivalence? What is the fundamental group $\pi_1(B)$? What is the fundamental group $\pi_1(C)$?
I would guess that $\pi_1(B)$ is the free group on uncountably many generators, since $B$ is the wedge sum of uncountably many circles.
For each $z \in S^1$, let $S_z = S^1$ be a copy of the circle, one for each point of $S^1$. Let $D = \bigsqcup_{z \in S^1} S_z$ be their disjoint union. Let $E = (S^1 \sqcup D)/\{\{z \in S^1, 1 \in S_z\} \mid z \in S^1\}$. This means that $E$ is obtained by gluing a point of each circle in $D$ to a point of $S^1$.
What is the fundamental group $\pi_1(E)$?
The space $B$ is indeed the wedge sum of uncountably many circles, hence its fundamental group is the free group on uncountably many generators, represented by the wedge summands. The space $C$ is obtained from $[0,1]$ by attaching $1$-cells $D^1_t,\,t\in[0,1]$ along constant maps $c_t\colon S^0_t\rightarrow[0,1],\,t\in[0,1]$. Thus, the pair $(C,[0,1])$ is cofibered, but $[0,1]$ is contractible, so the quotient map $C\rightarrow C/[0,1]=B$ is a homotopy-equivalence. In particular, $\pi_1(C)=\pi_1(B)$.
In a similar manner, the space $E$ is obtained from $S^1$ by attaching $1$-cells $D^1_z,\,z\in S^1$ along constant maps $c_z\colon S^0_z\rightarrow S^1,\,z\in S^1$. The homotopy type of $E$ (rel $S^1$) remains unaltered if we vary the attaching maps by a homotopy, so $E$ is homotopy-equivalent to the space obtained from $S^1$ by attaching $1$-cells $D^1_z,\,z\in S^1$ all along the same constant map $c_1\colon S^1_z\rightarrow S^1$, i.e. the wedge of (uncountably many plus one) circles, but this is of course abstractly homeomorphic to the wedge of uncountably many circles. Thus, $E\simeq B$ and $\pi_1(E)\cong\pi_1(B)$.