Suppose $X$ is a topological space with a cell decomposition, that is, a partition $\mathcal{E}$ of $X$ into open cells of various dimensions such that whenever $e \in \mathcal{E}$ is an $n$-cell, there exists a closed $n$-cell $D$ and a continuous map $\Phi : D \to X$, called a characteristic map, such that $\Phi$ restricts to a homeomorphism $\operatorname{Int}D \to e$ and maps $\partial D$ into the union of all cells in $\mathcal{E}$ of dimension strictly less than $n$.
When we specify a cell decomposition for $X$, we do not choose specific characteristic maps, leading me to think that they must be unique in some sense. Of course, characteristic maps are not unique, since we can precompose a given $\Phi$ with any homeomorphism $D \to D$ to obtain a different characteristic map.
But is it true that if $\Phi_1, \Phi_2 : D \to X$ are characteristic maps for $e \in \mathcal{E}$, then there exists a homeomorphism $f : D \to D$ such that $\phi_2 = \phi_1 \circ f$? I think we can find a counterexample if we take $X$ to be the line with two origins, but if we assume that $X$ is Hausdorff (as is traditionally done for cell complexes), does this uniqueness result hold?
No. Here is a counterexample. Consider $X = D^2$ with one $0$-cell $e^0 = \{1\}$, one $1$-cell $e^1 = S^1 \setminus e^0$ and one $2$-cell $e^2 = D^2 \setminus e^1$. As a charcteristic map $\Phi_1$ for $e^2$ we can take the identity on $D^2$.
Now let $S = \{ z \in S^1 \mid \Re (z) \ge 0\}$ be the closed right half-circle. It is well-known that the quotient space $D^2/S$ is homeomorphic to $D^2$. Let $p : D^2 \to D^2/S$ be the quotient map and $h : D^2/S \to D^2$ be a homeomorphism. Then also $\Phi_2 = h \circ p$ is a charcteristic map for $e^2$.