Retract of contractible space is contractible.
Let $r:X\to A$ where $A\subset X$: subspace and $r$ is retraction, $X$ is contractible.
Statement is so famous anybody can find the proof. But I found that many proofs is much longer than I thought. Here's my proof: $\text{id}_X\simeq C_{x_0}$ and $r\circ i = \text{id}_A$ where $i:A\hookrightarrow X$: inclusion. So, $\text{id}_A =r\circ i = r\circ\text{id}_X\circ i \simeq r\circ C_{x_0}\circ i=C_{r(x_0)}$ noting the fact that $f\simeq g\Rightarrow f\circ h\simeq g\circ h$ and $k\circ f\simeq k\circ g$ where each composition are well defined. I've never seen such proof in Google so far. They all use explicit homotopy $H:X\times I\to X$ from $\text{id}_X$ to $C_{x_0}$ so I wonder if my proof is wrong or have some serious problem.
Best Answer
Your proof is indeed correct. You rely on the critical notion that if $f \simeq g$ and $h \simeq k$ then $f \circ h \simeq g \circ k$. This observation is critical in another short proof of the fact.
Background:
Consider the "homotopy 1-category", where the objects are topological spaces and the arrows are continuous functions, quotiented by $\simeq$. The fact that this is actually a category is equivalent to the above observation about composition preserving $\simeq$ $\DeclareMathOperator{\id}{id}$
The terminal object of this category is the singleton space 1, because this is the terminal object of the category of topological spaces. In other words, there is only one map from a space $T$ to the singleton space 1, up to homotopy.
It can easily be seen that a space is contractible iff the space is isomorphic to 1 in the homotopy category.
If the space is isomorphic to 1, then consider the homotopy-isomorphism $f : 1 \to A$ and inverse $g : A \to 1$. Then we see that $\id_A \simeq f \circ g$, and $f \circ g$ is a constant function $C_{f(x)}$ where $x$ is the unique element $x \in 1$.
On the other hand, if $\id_A \simeq C_{a_0}$, then define $f : 1 \to A$ by $f(x) = a_0$ and let $g : A \to 1$ be the unique function. Then we see that $C_{a_0} = f \circ g \simeq \id_A$, so $A$ is homotopy-isomorphic to 1.
Thus, a contractible space is exactly a space which is a terminal object in the homotopy category.
End background
In general, if we have a terminal object $1$ and arrows $i : A \to 1$ and $r : 1 \to A$ such that $r \circ i = \id_A$, then $i$ is an isomorphism. This is because we have $i \circ r : 1 \to 1$ and $\id_1 : 1 \to 1$, and hence $i \circ r = \id_1$.
Thus, the retract of a contractible space is contractible.