# There exist no continuous one one onto (Bijective ) function form $[0,1]$ $\to$ $[0,1]\times[0,1]$ [duplicate]

I wanted to prove that there exist no continuous one to one and onto function form $[0,1] \to [0,1]\times[0,1]$.
My attempt : image of f on $[0,1]$ , a compact set is again compact .$[0,1]\times[0,1]$ is also compact.
On contrary suppose there exist continuous one one onto function between $[0,1]\to [0,1]\times[0,1]$ then its inverse function is also continuous one one onto.
Upto this I can write from given information
No idea how to proceed further .Any Help will be appreciated .

## marked as duplicate by Martin R, José Carlos Santos real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 5 '18 at 9:25

There is no such function because, since the domain is compact, it would be a homeomorphism. But $[0,1]$ and $[0,1]\times[0,1]$ are not homeomorphic: if you remove $\frac12$ from $[0,1]$, it becomes disconnected. But there is no such point in $[0,1]\times[0,1]$.