# On the existence of a continuous bijection $f\colon [0,1]\to [0,1]\times [0,1]$

Let $f$ be a continuous function on $[0,1]$ such that $f([0,1])=[0,1]\times[0,1].$ Then show that $f$ is not one-one.

Hints will be appreciated.

• You should look at Peano curve. – Nikita Evseev May 4 '15 at 10:53
• @NikitaEvseev That does not answer the question at all. – 5xum May 4 '15 at 10:54
• @Nikita Evseev. Thank you i looks some space filling curves. I am wondering to prove. – Mirunalini_UML May 4 '15 at 10:56
• helpful post – the.polo May 4 '15 at 11:11

Assume $f:K\rightarrow X$ where $K$ is a compact topological space, $X$ is a topological Hausdorff space and $f$ is continuous then for any closed set $F\subseteq K$ we have that $F$ is compact and hence $f(F)$ is compact in $X$ and hence closed in $X$.

Now if $f$ is one-to-one and onto this shows that $f^{-1}$ is continuous. And hence an homeomorphism.

Hint : apply this to your $f$ with $K=[0,1]$ and $X=[0,1]\times [0,1]$. You will get that $f$ is a homeomorphism between $K$ and $X$, finally you just have to show that it is not possible ($K-\{0.5\}$ and $X-\{f(0.5)\}$ should be homeomorphic as well).

• But is $f$ onto? If $f$ is a bijection, then we can use the result that continuous bijections from compact spaces to Hausdorff spaces are homeomorphisms, but I thought we were only given that $f$ is one-to-one? – Ilham May 4 '15 at 10:59
• @Clément Guérin I could not understand the first para. $F\subseteq X$ and $f(F) \subseteq X.$ Moreover, in any arbitrary topological spaces, Closed $\implies$ compact? – Mirunalini_UML May 4 '15 at 11:08
• @Iham, this a proof of the result you mention. – Clément Guérin May 4 '15 at 11:09
• @UmaMaheswariN.Lenin you are right. edited. – Clément Guérin May 4 '15 at 11:10
• @UmaMaheswariN.Lenin, I used the result (1) compact implies closed in $X$ (as Martini mentionned this is true only when the ambient space is Hausdorff), the result (2) closed implies compact in $K$ because $K$ is compact and (3) f(compact) is compact because $f$ is continuous. – Clément Guérin May 4 '15 at 11:12

Hint: For each $x\in [0,1]^2$,

$$[0,1]^2 - \{x\}$$ is connected, but

$[0,1] - \{1/2\}$ is not.

• Since connected is a topological property, $f$ can not be one-one. if it is, since domain is compact $f^{-1}$ exists and will be continuous. hence $f$ is homeomorphism. Thank You – Mirunalini_UML May 4 '15 at 11:16

Assume to the contrary that $f\colon [0,1]\rightarrow [0,1]\times [0,1]$ is a continuous bijection. Then its inverse is also continuous, i.e. it is a homeomorphism. Suppose that $f(\frac{1}{2})=(x,y)$.

Then $f$ induces a homeomorphism $g\colon [0,1]\setminus \{\frac{1}{2}\}\to ([0,1]\times[0,1])\setminus \{(x,y)\}$, but the domain is not connected, while the image is connected, which is a contradiction.

Hint: The space $[0,1]^2$ contains non-trivial closed jordan-curves. However, $[0,1]$ does not.

• Could you please explain that how to prove using this hint? Sorry i cant access the hint – Mirunalini_UML May 4 '15 at 14:45