Homotopy equivalence between O-O and $\theta$ Show that the dumbbell O-O (where there's no space between the "O" and "-") and the letter $\theta$ are homotopy equivalent, using the definition.
So, let $X$ be the set of points in the dumbbell, and $Y$ the set of points in $\theta$. We should give continuous maps $f:X\rightarrow Y$ and $g:Y\rightarrow X$ such that $f\circ g$ is homotopic to $\text{id}_Y$ and $g\circ f$ is homotopic to $\text{id}_X$.
I'm thinking about mapping the "-" in the dumbbell to the $-$ in $\theta$, and the two "O"s in the dumbbell to the two halves of $\theta$. But each end of the "-" in the dumbbell is connected to only one half, while each end of the $-$ in theta is connected to both halves. So I'm not sure what to do.
 A: Let $v$ be a point in the middle of the the central bar of O-O and $w$ a point in the central bar of $\theta$.
Map the central bar of O-O to $w$, and the left and right loops in O-O to the upper and lower loops based at $w$ in $\theta$, maintaining orientation.
Similarly, map the central bar in $\theta$ to $v$, and map the two semicircles in $\theta$ to the two loops in O-O based at $v$, maintaining orientation.
In both directions, the composition uniformly "pulls in" the bar into the central vertex and pulls in some of the loops into the bar, so they are homotopic to the identity.

A: Further to Léo's comment, I will use an intermediate step; consider the unit disk $D^2$ with two smaller disks removed. I will show the three graphs (figure eight, dumbell, theta) will all be homotopy equivalent to this space. Writing $B(\varepsilon, x)$ for the open ball of radius $\varepsilon$ based at $x$, define
$$D=D^2-B(1/4,1/2)-B(1/4,-1/2).$$
Notice that $D$ deformation retracts onto the three aforementioned graphs, $G$, if we position them appropriately (i.e. as subsets of $D$ with the points $\pm1/2$ inside the "loops" of the dumbell, figure eight, and theta). So there exists a homotopy $f_t\colon D\to D$ such that $f_0=id_D$, $f_1(D)=G$, and $f_t\restriction_G=G$ for all $t$. Let $i\colon G\to D$ be the inclusion; now, $f_1\circ i=id_G$, and $i\circ f_1^\prime =f_1$ (where $f_1^\prime$ is $f_1$ viewed as a map $f_1\colon D\to G$). Of course $id_G\simeq id_G$, and $f_1\simeq id_D$ by the homotopy $f_t$. 
All three graphs are homotopy equivalent to $D$, and hence to eachother by transitivity of $\simeq$.
