Here is the question:
Let $X$ be a normal space and let $A \subseteq X$ be an arc, with inclusion function $i: A \rightarrow X,$ Since $A$ is an arc, we have a homeomorphism $\bar{\alpha} : I \xrightarrow{\cong} A;$ Let's write $f_{A} =\bar{\alpha}^{-1}: A \rightarrow I.$ Prove the following "Let $X$ be a normal space. Then every arc $A$ in $X$ is a retract of $X$." by analyzing the following diagram:
$$\require{AMScd} \begin{CD} A @>{f_{A}}>> I\\ @VVV @VVV \\ X @>{f}>> I \end{CD} \longrightarrow A$$
Where the first vertical arrow is the inclusion map, the second vertical arrow should not exist but I am not skillful in drawing commutative diagrams and the last horizontal arrow on the right (that goes into $A$) should start from $I.$ and the arrow that has the function $f$ above it should be dotted line because we are searching for this function $f.$
My thoughts:
I know that:
1- -A space $X$ is said to be normal if all singleton sets are closed and if for any two disjoint closed subsets $A, B\subseteq X$ there are open sets $U,V \subseteq X$ such that $$A \subseteq U, B \subseteq V, \textbf{ and } U \cap V = \emptyset.$$
2-Definition
An arc in a topological space $X$ is a subspace $A \subseteq X$ that is the image of an injective continuous function $\alpha : [a,b] \rightarrow X$ from a closed interval in $\mathbb{R}$ to $X.$ If $X$ is Hausdorff, then the map $\bar{\alpha} : [a,b] \rightarrow A$ obtained from $\alpha$ by restricting the target is a homeomorphism.
The homeomorphism $\bar{\alpha}$ enables us to identify the two endpoints of an arc $A \subseteq X-$ they are the points $\bar{\alpha}(a)$ and $\bar{\alpha}(b).$
Arcs differ from paths in 2 ways:
1- Arcs are subsets, not functions.
2- Arcs are not even just images of paths: an arc must be homeomorphic to $[a,b],$ while the image of a generic path can be very different from $[a,b].$
3- I am allowed to use "Tietze Extension Theorem" statement without proving it. So by it I am given that $f \circ i = f_{A}.$ Here is the statement I am allowed to use:
Let $X$ be a normal space and let $A \subseteq X$ be a closed subspace with the inclusion map $i: A \rightarrow X.$ If $J \subseteq R$ is any closed interval and $ f_{A}: A \rightarrow J$ is any continuous function, then there exists a continuous function $f: X \rightarrow J$ making the following diagram commutes:
$$\require{AMScd} \begin{CD} A @>{f_{A}}>> J\\ @VVV @VVV \\ X @>{f}>> J \end{CD} $$
Where the first vertical arrow is the inclusion map, the second vertical arrow should not exist but I am not skillful in drawing commutative diagrams. and the arrow that has the function $f$ above it should be dotted line because we are searching for this function $f.$
4- To prove that every arc $A$ is a retract of $X,$ I know that I have to prove that there exists a continuous function $r: X \rightarrow A$ such that $r\circ i = id_{A}$ where $i:A\rightarrow X$ is the inclusion map.\
My idea is:
Since $X$ is normal, $A$ is an arc, $I$ is a closed interval and since $\bar \alpha$ is a homeomorphism then $f_{A}$ exists and is continuous and then by Tietze extension theorem there exists a continuous function $\bar{f}_{A}: X \rightarrow I$ such that $f = \bar{f}_{A} \circ i = f_{A}.$ then if I compose on both sides with $\bar \alpha,$ I will get that the RHS is the $id_{A}$ but then the LHS will be $\bar{f}_{A} \circ i \circ \bar \alpha.$ I need to take my retraction $r$ to be $\bar{f}_{A} \circ \bar \alpha$ but how can I do that? The inclusion function $i$ is inbetween. Could anyone help me in solving this situation please?
5- I understood from this link A retract in normal space that Urysohn's Lemma will be used but I do not understand why we should use it? Also my version of Tietze theorem does not include extention of the codomain. So how will my proof be different from the one I mentioned in the above link?
Since $I$ is a retract of $\mathbb R$, with retract $r:\mathbb R \rightarrow I$ we can always extend a function $f: A \rightarrow I$ to $X \rightarrow I$. Just extend $f$ to $\tilde f: X \rightarrow \mathbb R$ and then $r \circ \tilde f$ extends $f$ with the appropriate codomain.
Your retract $X \rightarrow A$ is constructed using the extension of $f_A: A \rightarrow I$, denoted $g:X \rightarrow I$, then $\tilde \alpha \circ g$ will be the retract $X \rightarrow A$ you are looking for. This is because $(\bar \alpha \circ g) \circ i = \bar \alpha \circ f_A = \text{id}_A$ by definition so $\bar \alpha \circ g$ really is a retract $X \rightarrow A$.