I am learning Algebraic Topology. While studying the definition of Homotopy Equivalence I thought myself to apply this concept in the case of some "easy" sets such as $[0,1]$ and $(0,1)$. This would help me understand the concept better.
So let $f:[0,1] \to (0,1)$ be defined by $f(x)=\frac {x+1}{3}$ and $g:(0,1) \to [0,1]$ be the inclusion map i.e. $g(x)=x \; \forall \; x \in (0,1)$. Then $f$ and $g$ both are continuous.
First we show that $f \circ g \simeq Id_{(0,1)}.$
Define $H:(0,1) \times [0,1] \to (0,1)$ such that $H(x,t)=tx+(1-t)\frac {x+1}{3}$. Then $H(x,0)=\frac {x+1}{3}$ and $H(x,1)= x \; \forall \; x \in (0,1).$ This is the required homotopy.
Next we show that $g \circ f \simeq Id_{[0,1]}.$
Define $H:[0,1] \times [0,1] \to [0,1]$ by $H(x,t)=tx+(1-t)\frac {x+1}{3}$. Then $H(x,0)=\frac {x+1}{3}$ and $H(x,1)= x \; \forall \; x \in [0,1].$ This is the required homotopy.
Thus $[0,1]$ and $(0,1)$ have same homotopy type i.e. they are homotopy equivalent.
Is my proof and hence the understanding of the concept correct here? Thanks in advance.
Yes, this is correct.
A slightly easier way to go about it is to note that both are contractible, meaning they have the homotopy type of a point. To show this, you just have to find homotopies from the identity maps $[0,1]\rightarrow [0,1]$ and $(0,1)\rightarrow(0,1)$ to constant maps, which you can do with the same intuition you used for your own proof.
In general, if you can find a deformation retract of a space $X$ onto a subspace $Y\subseteq X$, then $Y$ and $X$ are homotopy equivalent. This tends to give easier geometric reasoning, as we can then interpret your proof as showing that you can "contract" the ends of $(0,1)$ to get a deformation retract onto the interval $[1/3,2/3]$, which is of course homeomorphic to $[0,1]$.