Show that for two vector spaces there exist a surjective linear map $\varphi$ 
If $V = \{(x,y,z) \in \mathbb{R^3} : x+2y-z=0 \}$ and $W=\{(x,y,z) \in \mathbb{R^3} : -x+z=0 \}$ then show that there exists a surjective linear map $\varphi: V\to W$.

How can I show this when $\varphi$ is not given? $\varphi$ is surjective iff $\operatorname{im}(\varphi)=W.$ This is just to show that two sets are the same so i'm thinking of proving this by showing $\operatorname{im}(\varphi) \subset W$ and $W \subset \operatorname{im}(\varphi)$, but i cannot do it. Is this the right way?
 A: Both spaces have dimension 2, since each equation (hyperplane) cuts out one dimension of the ambient space.
Here $\{(1,0,-1),(2,-1,0)\}$ is a basis of $V$ and $\{(1,0,1),(0,1,0)\}$ is a basis of $W$.
Any mapping $V\rightarrow W$ which provides an assignment of the basis vectors is linear, such as $(1,0,-1)\mapsto (1,0,1)$ and $(2,-1,0)\mapsto (0,1,0)$, and in this case also surjective.
A: *

*$V\cong\Bbb R^2$ since any vector in $V$ may be written $$(x,y,z)=x(1,0,1)+y(0,1,2)$$
(any vector may be written as a linear combination of two vectors, which themselves are linearly independent. Therefore $\dim_\Bbb R(V)=2$)

*Also, $W\cong\Bbb R^2$ since again any vector in $W$ may be written $$(x,y,z)=x(1,0,1)+y(0,1,0)$$
Therefore $V\cong W$, so there is an isomorphism $\varphi: V\rightarrow W$, in particular $\varphi$ is surjective.

You can always find a surjective linear map $\varphi:V\rightarrow W$ if $\dim(W)\leq\dim(V)$. Just map a basis in $V$ to basis vectors in $W$. Let $\mathfrak B_V=\{v_1,v_2,\ldots,v_n\}$, $\mathfrak B_W=\{w_1,w_2,\ldots,w_m\}$ be a basis for $V$ and $W$ respectively. Then consider the map $$\varphi:V\rightarrow W\\\varphi(v_i)=w_i$$ To see that this map is surjective let $a_1w_1+\ldots+a_mw_m\in W$ be any vector, notice that $$a_1w_1+\ldots+a_mw_m=a_1\varphi(v_1)+\ldots+a_m\varphi(w_m)\\=\varphi(a_1v_1+\ldots+a_mv_m)$$ by definition of a linear map. Thus, you have the desired vector $a_1v_1+\ldots+a_mv_m$.
