Remark in Hatcher's Algebraic Topology on Mapping Cylinders On p2 of Algebraic Topology, Hatcher defines mapping cylinders as follows:

For a map $f: X \to\ Y$, the mapping cylinder $M_f$ is the quotient space of the disjoint union $(X \times I) \cup Y$ obtained by identifying each $(x,1) \in X \times I$ with $f(x) \in Y$.

Then, he makes the following remark:

Not all deformation retractions arise in this simple way from mapping cylinders. For example, the thick $\bf{X}$ deformation retracts to the thin $X$, which in turn deformation retracts to the point of intersection of its two crossbars. The net result is a deformation retraction of $\bf{X}$ onto a point, during which certain pairs of points follow paths that merge before reaching their final destination.

I'm unclear as to why the deformation retraction described in the paragraph above does not arise from a mapping cylinder. Hatcher seems to hint that the merging of paths is the reason why, but I do not see how merging of paths is inconsistent with the definition of a mapping cylinder.
Any thoughts?
 A: I am not entirely sure how to make it perfectly formal, but here are my thoughts.
In the letter $\mathbf X$ example, the space $X$ that would serve in our definition of the mapping cylinder is the outline of the thick $\mathbf X$, which is homeomorphic to $S^1$. The space $Y = \{P\}$ consisting of the point $P$ at the center of the $\mathbf X$ serves as the codomain in our mapping cylinder. Thus our mapping cylinder $M_f$ is the quotient of $(S^1\times [0,1])\sqcup Y$ obtained by identifying each point of $S^1\times\{1\}$ with $P$, and it is easy enough to see that the mapping cylinder is thus homeomorphic to a closed disk $D^2$.
Now I think the point of what Hatcher is getting at is that the overall deformation retraction $\{\tilde f_t\}$ consisting of the composition of first turning $\mathbf X$ to the thin $\mathsf X$, and then the thin $\mathsf X$ to the point $P$ is more complicated/has more information than the mapping cylinder of the map $f$ that sends all of $\mathbf X$ to $P$. If for each $t\in[0,1]$, you imagine the map $f_t\colon X\to (M_f)_t$ to be the map that sends $X$ to the horizontal $t$-slice $(M_f)_t$ of the mapping cylinder $M_f$, then together the family $\{f_t\}$ is not the same as $\{\tilde f_t\}$. Further, I think that you can make "more complicated/has more information" more precise by noting that for all $0\le t < 1$, the map $f_t$ is a homeomorphism, but the same is not true for the family $\{\tilde f_t\}$.
