I have constructed the following proof of the statement and have some questions (a question) about the correctness of the proof:
Statement:
$A$ retract of $X$ and $X$ contractible implies $A$ contractible.
Proof:
I make the following observations:
- Since $X$ is contractible we have $1_{X} \simeq constant $. This follows from a proposition that a space $X$ is contractible if and only if $1_{X}$ is nullhomotopic.
- $A$ retract of $X$ gives us that there exists a map $r: X \rightarrow A$ such that $r\circ i = 1_{A}$ where $i:A\rightarrow X$ is the inclusion map.
Now I conclude:
$1_{A} = r\circ i = r \circ 1_{X} \circ i \simeq r \circ constant \circ i$
Since $r \circ constant \circ i$ is a constant map we have that $1_{A}$ is nullhomotopic and thus we have $A$ is contractible.
Questions:
- Is this proof correct?
- I use that $f \simeq f_{1}$ implies $g \circ f \simeq g\circ f_{1}$. Is this true? I think it makes sense.