Let $X \subset \Bbb R^2$ be a union of the coordinate axes and the line $x+y=1$, $0\le x \le1$. Show that $X$ is homotopy equivalent to $\Bbb S^1$. 
Let $X \subset \Bbb R^2$ be a union of the coordinate axes and the line $x+y=1$, $0\le x \le1$. Show that $X$ is homotopy equivalent to $\Bbb S^1$.

Denote the triangle formed by $(0,0),(1,0),(0,1)$ as $K$. The trick here is apparently to show that $K \simeq X$ and then using the fact that $K$ is homeomorphic to $\Bbb S^1$ to deduce that $X \simeq \Bbb S^1$.
With this I've managed to get the following. Define $f :X \to K$ as $$f(x) = \begin{cases} x, & x \in K \\ (0,1), &x \in \{0\} \times [1,\infty) \\ (1,0), & x \in [1,\infty) \times \{0\} \\ (0,0), & x\in \{0\} \times (-\infty, 0] \\ (0,0), &x \in (-\infty, 0] \times \{0\} \end{cases}$$ and define the inclusion $\iota :K \to X$. We know have that $f \circ \iota = id_K$ and I think I can define $h:K \times [0,1] \to X$ as $$h(a,t)=(1-t)(\iota \circ f)(a) + t \cdot id_X(a)$$ to show that $\iota \circ f \simeq id_X$?
The problem I'm having is that I didn't know that $K$ is homeomorphic to $\Bbb S^1$. What is the map giving this homeomorphism?
 A: Denote by $T$ the aforementioned triangle.
For $f:T\to X$, just use inclusion.
For $g:X\to T$, you know that each vertex $v$ of the triangle is the meeting point of two lines $L,M$? Now, $L-\{v\}$ has two components. One includes the edge of the triangle, the other doesn't. It would be a good idea to map every element of the latter component to the vertex $v$ -- and to do likewise for all lines and vertices. The process is easier to understand if you draw it on a piece of paper.
That is one example of $g$ that would be enough to show homotopic equivalence.
$f\circ g$ is the identity on $X$ itself.
To show that $g\circ f$ is homotopic to the identity on $Y$, take each half-line that has been collapsed into a vertex, and pull it back out.
A: To prove that $K$ is homeomorphic to $\mathbb S^1$, use any circle $C$ contained in the solid triangle bounded by $K$ (such as the inscribed triangle). You already know that any two circles in the plane are homeomorphic, so what remains is to construct a homeomorphism $f : K \to C$. Let $p$ be the center and $r$ the radius of $C$. For each $q \in K$, visualize the ray based at $p$ and passing through $q$, and let $f(q) \in C$ be the unique point where that ray passes through $C$. From this you will obtain the formula
$$f(q) = p + r \cdot \frac{q-p}{\|q-p\|}
$$
Prove that $f : K \to C$ is a continuous bijection, and then use compactness of $K$ to prove that $f$ has continuous inverse.
