Finding a homotopy equivalence between the letters $X$ and $Y$ I'm trying to show that the letters $X$ and $Y$ are homotopy equivalent but I'm having some trouble.
If I let $f$ map three half open prongs of the $X$ into the $Y$ and the last closed prong into the center of the $Y$, and let $g$ map three half open prongs of the $Y$ into the half open prongs of the $X$ with the center of the $Y$ being mapped to the center of the $X$, what would be a homotopy between $g \circ f$ and $Id_{X}$ and a homotopy between $f \circ g$ and $Id_{Y}$?
 A: You can simplify a bit the situation by supposing that the letter $Y$ is a subspace of $X$. Then you get a first natural map $g:Y\to X$ which is just the inclusion.
So now, you need to push the "right leg" of $X$ to the center. Let $f:X\to Y$ be that map, such as $f|_Y$ is just the identity.
To conclude, you first need to prove that $f\circ g : Y\to X\to Y$ is homotopic to identity of $Y$, but that's clear since $f|_Y$ is the identity. Secondly you need $g\circ f : X\to Y \to X$ to be homotopic to the identity of $X$.
That's not very difficult. Basically you need to see why this homotopy is quite the same as the one which transport the path $\gamma(t)=t$ to $\delta(t) =0$.
A: To have concrete spaces to work with, let $X = \{ (x,y) \in \mathbb{R}^2 : |x| \leq 1 \text{ and } y=\pm x \}$ as a subspace of the plane. This is just the superposition of the graphs of the lines $y = \pm x$ restricted to the interval $[-1,1]$. Let us consider $Y = X \setminus \{ (x,-x) : x > 0 \}$ as a subspace of $X$ (this is just $X$ with the lower right leg removed). We have the inclusion $i: Y \to X$. The other way around, we have a map (called a retraction) $r:X \to Y$ defined by
$$ r(x,y) = \begin{cases}(x,y) & \text{if } (x,y) \in Y \\ (0,0) &\text{otherwise}
\end{cases}.$$
The idea is that $r$ pushes the lower right leg of $X$ into the center of the figure. However, it does so in one fell swoop. In order to turn this into a continuous motion, which slowly pushes the lower right leg into $X$, we need to provide homotopies. We have $ri = 1_Y$ directly, so the constant homotopy suffices. However it is not true that $ir = 1_X$, because points in the lower right leg are mapped to the center and never return to their original position. We may construct a homotopy $ir \simeq_h 1_X$ as follows: $h:X \times I \to X$ is defined by $$h((x,y),t) = \begin{cases}(x,y) & \text{if } (x,y) \in Y \\ t(x,y) &\text{otherwise} \end{cases}.$$
This homotopy starts at $ir$ and ends at the identity $1_X$. Therefore we have shown that $X$ and $Y$ have the same homotopy type. (Provided we check that all the given maps are continuous, which I'll leave to you as an exercise. One uses the pasting lemma to justify the continuity of piecewise-defined maps such as $h$.)

Note that it would be even easier, or at least more satisfying, to just show that both spaces are contractible, i.e., homotopy equivalent to a point. To do so, we may use the same kind of argument, by showing that the inclusion of the center point and the retraction of both figures to the center point are homotopy inverses. In this case the homotopy from $ir$ to $1_X$ or $1_Y$ takes the simple form $h((x,y),t) = t(x,y)$, and this works for both $X$ and $Y$.
