# Existence of a continuous function.

Question is to check Which of the following statements are true?

1. There exists a continuous function $f: \{(x,y)\in \mathbb{R}^2 : 2x^2+3y^2=1\}\rightarrow \mathbb{R}$ which is one-one.
2. There exists a continuous function $f: (-1,1)\rightarrow (-1,1]$ which is one one and onto.
3. There exists a continuous function $f: \{(x,y)\in \mathbb{R}^2 : y^2=4x\}\rightarrow \mathbb{R}$ which is one-one.
4. There exists a continuous function $f: \{(x,y)\in \mathbb{R}^2 : x^2+y^2=1\}\rightarrow \mathbb{R}$ which is Onto.
5. There exists a continuous function $f : S^1=\{(x,y)\in \mathbb{R}^2 : x^2+y^2=1\} \to \mathbb{R}$ which is one-one.

What i have done so far is :

Solution for $1$ :

we see that $\{(x,y)\in \mathbb{R}^2 : 2x^2+3y^2=1\}$ is connected and compact...

Suppose we have continuous function $f :S \rightarrow \mathbb{R}$ which is one one then we would see that $S$ is homeomorphic to $f(S)$*

Now, $S$ is connected so is $f(S)$ and $S$ is compact so is $f(S)$.Thus $f(S)=[a,b]$ for some $a,b\in \mathbb{R}$

suppose I remove a point $c\in(a,b)$ then $f(S)$ would have two connected components where as the corresponding subset of $S$ obtained by removing $f^{-1}(c)$ is connected which is absurd. Thus there exist no continuous one one map $f :S \rightarrow \mathbb{R}$

Solution for $2$ :

Suppose we have continuous bijection $f: (-1,1)\rightarrow (-1,1]$ then we have $t\in(-1,1)$ such that $f(t)=1$.

Now, any continuous injection has to be such that $a<b$ implies $f(a)\leq f(b)$ or $f(a)\geq f(b)$ for all $a,b$ (I do not how does one call this property as)

Assuming $f$ is increasing and $f(t)=1$ then $f(m)\geq 1$ for all $m\in (t,1)$ which is a contradiction to $f$ being injective...

Assuming $f$ is decreasing and $f(t)=1$ then $f(m)\geq 1$ for all $m\in (-1,t)$ which is a contradiction to $f$ being injective...

Thus there is no continuous bijection from $(-1,1)$ to $(-1,1]$

Solution for $3$ :

we see that $\{(x,y)\in \mathbb{R}^2 : y^2=4x\}$ is connected (Not Compact. sorry for my laziness and thanks to gaoxinge :D )...

Suppose we have continuous function $f :S \rightarrow \mathbb{R}$ which is one one then we would see that $S$ is homeomorphic to $f(S)$*

Now, $S$ is connected so is $f(S)$ and $S$ is compact so is $f(S)$.Thus $f(S)=[a,b]$ for some $a,b\in \mathbb{R}$

I though of applying same idea as i have done to check First question but then i am not getting any negative result.

Moreover I guess that such a map exists and i can actually find a homeomorphism to $\mathbb{R}$.

I am unable to write explicitly but What i would do is i would bend the parabola so that it would coincide with real line (I am not able to express it precisely please try to understand something from this). But how would i write "bending" as a function and how would i show that this is a homeomorphism. Please help in that case and more over is my intuition correct?

Solution for $4$ :

Continuous image of compact space is compact but $\mathbb{R}=f(S)=f(\{(x,y)\in \mathbb{R}^2 : x^2+y^2=1\})$ is not compact though $S=\{(x,y)\in \mathbb{R}^2 : x^2+y^2=1\}$ is compact.

Solution for $5$ :

for the same reason that i have treid to explain in First question there is no continuous function $f : S^1=\{(x,y)\in \mathbb{R}^2 : x^2+y^2=1\} \to \mathbb{R}$ which is one-one.

Please see if this justification is sufficient and if it is please let me know if there are any other ways of seeing these things.

Thank you.

P.S: I have used one result which i feel is worth sharing :

continuous bijection from a compact space to Hausdorff space is homeomorphism

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(1) and (5) is talking about that is $\mathbb S^1$ homemorphic to $\mathbb R^1$? You use the compactness to judge they are different. It's awesome! Also you can compute them topological invariant like fundamental group $\pi_1$.
(2) is related to the open and closed property and you use the monotone property to judge. It's very well!! Actually, $(0,1)$ is an one dimensional manifold without boundary and $(0,1]$ is an one one dimensional manifold with boundary.
(3) I think you say $\{(x,y):y^2=4x\}$ is compactness it's wrong. Actually, it is parabolic curve which is closed but not bounded. And you say you find the homemorphism is right.