1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Thomas Andrews Oct 18 '17 at 21:16
  • $\begingroup$ What is $A\phantom{}$? $\endgroup$ – anomaly Oct 18 '17 at 21:18
  • $\begingroup$ Ah good point. I'll change that. $\endgroup$ – Sid Caroline Oct 18 '17 at 21:18
  • $\begingroup$ $g(\overline x)=0$ 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. $\endgroup$ – Thomas Andrews Oct 18 '17 at 21:22
  • 1
    $\begingroup$ Do you know the Tietze extension theorem? $\endgroup$ – Eric Wofsey Oct 18 '17 at 21:47
1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ So if I use Tietze extension theorem, the proof is essentially a one-liner? $\endgroup$ – Sid Caroline Oct 18 '17 at 22:42
  • $\begingroup$ 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). $\endgroup$ – JonathanZ Oct 19 '17 at 0:48
  • $\begingroup$ 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 $[0,1]^A$ $\endgroup$ – JonathanZ Oct 19 '17 at 0:58
  • $\begingroup$ I don't think it requires $[0,1]^A$ to be normal. $\endgroup$ – Sid Caroline Oct 19 '17 at 2:28
  • 1
    $\begingroup$ Normality is one of the conditions required to apply Tietze, at least as it is stated in the Wikipedia page. $\endgroup$ – JonathanZ Oct 19 '17 at 2:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.