# Can a “dimension reducing” function be bijective?

Can a function whose image has lower dimensionality than its domain (alas I don't know if there is a special name for that kind of functions) ever be bijective? Consider e. g. $$f:X\to Y:\begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix} \to\begin{pmatrix} y_1\\y_2 \end{pmatrix}=\begin{pmatrix} x_1 +x_2\\x_2+x_3 \end{pmatrix}$$ which is a function from $$\Bbb{R}^3$$ into $$\Bbb{R}^2$$ and is obviously not bijective. My problem is: On the one hand, I think it should be possible to have a bijective function from a "cube" into a "plane", on the other, it's hard to imagine because it effectively means calculating one result variable from multiple input variables, which would imply that the same image element can always result from different domain elements, thus rendering bijectivity impossible.

• The existence of a bijection between two sets is equivalent to the sets having the same cardinality, and it is well known that objects of different dimension (I use dimension in the sense that you used it) can have the same cardinality, i.e. take the unit interval and unit square. So you can form a bijection between the unit interval and the unit square, but clearly they have different dimensions. Edit: Just saw that 5xum already mentioned this example – StannisBa May 10 at 12:18
• Any such bijection would not be continuous, but very non continuous examples of such functions exist. – Somos May 10 at 19:12

$$f(0.a_1a_2a_3a_4,\dots, 0.b_1b_2b_3\dots) = 0.a_1b_1a_2b_2a_3b_3a_4b_4\dots$$
which can be well defined with correct handling of some edge cases involving infinite nines, and it can be shown it is a bijection that maps $$(0,1)^2$$ to $$(0,1)$$.
If $$X$$ is an infinite set then $$X^n$$ will have the same cardinality for finite $$n$$. Hence you can find bijections between any pair. 5xum gives an example of how to do this with $$\mathbb{R}^2$$ to $$\mathbb{R}$$.
However, if you want the map to have some nice properties, e.g. linear, then it might not be. If $$U$$ and $$V$$ are vector spaces and $$f: U \rightarrow V$$ is a linear map and a bijection then $$U$$ and $$V$$ will have the same dimension.