TITLE: Does there exist a continuous bijective function $ \displaystyle f : \Bbb R \rightarrow \Bbb R − \{1\}$?
QUESTION [8 upvotes]: This is my first time posting on here :)
My question is this: Does there exist a continuous bijective function $f :\Bbb R → \Bbb R − \{1\}$? Explain yes or no...
My thoughts:
Let $ \displaystyle f:  A \rightarrow  B$ be a function 
Okay for there to be a bijective function, for every element in A, there must be a unique element mapping $A$ to $B$. Right? Well if that is the case, then the answer to the question is FALSE. Why? Because the cadinality of the domain of $A$ (which in this case is R) is GREATER than the cardinality of the co-domain which is ($\Bbb R-\{1\}$). 
Because this is so, there must be some duplicates such that two different elements in $A$ map to the same element in $B$.
Does this make sense?
Unfortunately, I would be happy with this answer but another question pops up in my mind. I have read in my textbook that the cardinality of $\Bbb N$ is equal to the cardinality of $\mathbb Z$  
But how? Since $\Bbb N$ represents all natural numbers ($0,1,2,3...$) and $\mathbb Z$ represents all the integers ($...-3,-2,-1,0,1,2,3...$). Clearly the cardinality of $\mathbb Z$ is greater than N but still the two have equal cardinality. 
So tying that in with the question, my answer doesn't seem right anymore.
Any guidence or help would be greatly appreciated!!!
Thanks :)

REPLY [6 votes]: You can't do this by cardinality arguments, since $\mathbb{R}$ and $\mathbb{R}\setminus\{1\}$ actually have the same cardinality. You need some topological argument.
The keyword is connectedness. The continuous image of a connected set like $\mathbb{R}$ must be connected, unlike $\mathbb{R}\setminus\{1\}$. So your function can't even be continuous and onto. 
Even though it boils down to the same in $\mathbb{R}$, it is slightly easier to consider convexity. Recall a set $S$ is convex in an affine space if, for every $s_1,s_2$ in $S$, and every $t\in [0,1]$, the element $(1-t)s_1+ts_2$ belongs to $S$. That is, $[s_1,s_2]\subseteq S$ as soon as $s_1,s_2$ belong to $S$.
By the intermediate value theorem, the continuous image of a convex set in $\mathbb{R}$ is convex. Now $\mathbb{R}$ is convex, but $\mathbb{R}\setminus\{1\}$ is not. So there does not even exist a continuous function from $\mathbb{R}$ onto $\mathbb{R}\setminus\{1\}$.