If $f:[0,1]\rightarrow [0,1]^A$ is an embedding

Exercise: Given a continuous injection $f:[0,1]\rightarrow [0,1]^A$, there exists a continuous function $g:[0,1]^A\rightarrow [0,1]$ such that $g\circ f$ is the identity map.

Edit:

If I can show that $f([0,1])$ is closed in $[0,1]^A$, then we can define $g:f([0,1])\rightarrow [0,1]$ by sending $\bar{x}\in f([0,1])$ to $f^{-1}(\bar{x})$. $g$ is well defined since $f$ is bijective; $g\circ f$ is the identity map; and $g$ is continuous (but why?).

By Tieze extension theorem we can extend $g$ to a continuous function $g':[0,1]^A\rightarrow [0,1]$, and $g'\circ f$ is the identity map.

$g$ is essentially $f^{-1}$ restricted to $f([0,1])$, and $f:[0,1]\rightarrow f([0,1])$ is continuous and bijective. But why is $g$ also continuous?

Since $[0,1]$ is a compact space and $f$ is continuous, $f([0,1])$ must also be compact in $[0,1]^A$. Also $[0,1]$ is Hausdorff, so $[0,1]^A$ is Hausdorff. So $f([0,1])$ is a compact subset of a Hausdorff space $[0,1]^A$, hence it must be closed.

• Wait, what does $g(\overline x):=[0,1]$ mean? A function is single valued, and its value is an element of the range, not the whole range. – Thomas Andrews Oct 18 '17 at 21:16
• What is $A\phantom{}$? – anomaly Oct 18 '17 at 21:18
• Ah good point. I'll change that. – Sid Caroline Oct 18 '17 at 21:18
• $g(\overline x)=0$ isn't going to be continuous, since for any $a\neq 0$ any neighborhood of $f(a)$ will have points that are not in the image of $f$ (at least if $|A|>1$.) So that won't be continuous. – Thomas Andrews Oct 18 '17 at 21:22
• Do you know the Tietze extension theorem? – Eric Wofsey Oct 18 '17 at 21:47

I'm pretty sure @Thomas Andrews is right, that it won't be continuous. Try sketching out the simple example of $f : [0,1] \rightarrow [0,1]^2, x \mapsto (x,x)$ to get an idea. (I don't see that there is any fix, but I haven't thought about it for very long.)
Also, if you want to work out the specific details for this example to develop your understanding then go ahead, but I also think that (extending @Eric Wofsey's comment) that you can prove a much more general version of your claim just using the facts that $[0,1]$ is compact and $[0,1]^A$ is normal (thus allowing you to use Tietze).
• Yes, basically -- maybe let's say "two to three liner". You have to establish that the inverse of $f$ (with its domain restricted to the image of $f$) is continuous (this uses compactness), and then you have to establish that $[0,1]^A$ is normal (confession: I don't know that it is, but most "nice enough" spaces are normal so I'm betting you can find a basic theorem on normality that will say it is). – JonathanZ Oct 19 '17 at 0:48
• Oh, I missed your edit to your question. Yes, that's exactly the idea I was thinking of, although you're missing the bit about establishing normality of $[0,1]^A$ – JonathanZ Oct 19 '17 at 0:58
• I don't think it requires $[0,1]^A$ to be normal. – Sid Caroline Oct 19 '17 at 2:28