Find that the given linear transform is a isomorphism I'm studying Linear Algebra and I'm having trouble demonstrating that a function is a isomorphism, that is:
 "Given the linear transform $T: V \rightarrow W$, $T$ is a isomorphism if and only if it is both injective and surjective."
The function that I'm analyzing is $T(x, y, z) = (x-2y, z, x+y)$. I think it manged to show that the function is injective as the following demonstration:
If $T$ is injective, then $f(x) = f(y) \implies (x) = (y)$. Therefore:
$$T(x, y, z) = x(1, 0,1) + y(-2, 0, 1) + z(0,1,0) = a(1,0,1) + b(-2, 0, 1) + c(0,1,0) = T(a, b, c)$$
But I'm having trouble showing that said function is surjective. If I'm not wrong a surjective function is one that: Given the domain $X$ and co-domain $Y$, $\forall y  \in Y, \exists  x \in X $ such as $f(x) = y $
 A: You proof of injectivity isn't clear (at least to me).
You assume that $T(x,y,x) = T(a,b,c)$. This implies
$$
(x - 2y, z, x+y) = (a - 2b, c, a + b).
$$
Then it is clear that $z = c$. Now you just have to show that
$$
x - 2y = a-2b \\
x+y = a+b
$$
implies that $x = a$ and $y = b$.
If $T$ is a map from $\mathbb{R}^3$ to $\mathbb{R}^3$ and it is injective, then it is automatically surjective. This is because the image of a surjective map has dimension equal to the domain.
If you are not happy with saying this, you can, of course, show surjectivity by saying that given $(a,b,c) \in \mathbb{R}^3$ the system of equations
$$\begin{align}
x - 2y &= a \\
z &= b \\
x + y &= c
\end{align}
$$
has a solution. That is, you show that given $(a,b,c)$ there is $(x,y,z)$ such that $T(x,y,z) = (a,b,c)$.
A: To show that your linear map is surjective, you just have to check that there exists a set of elements in the domain which is mapped to a spanning set of codomain.
In particular: $T(0,0,1)=(0,1,0)$, $T(1,0,0)=(1,0,1)$ and $T(0,1,0)=(-2,0,1)$
then check that above three vectors are linearly independent, and hence must be a basis your codomain. Then by linearity of the mapping, any element in your codomain must be expressed in terms of these vectors, and hence T is surjective.
Also, your argument for injectivity is slightly wrong in a sense that: You stated that $T(x,y,z)=T(a,b,c)$ but you did not say anything about $(a,b,c)$ and $(x,y,z)$, although it is clear then this implies that $(a,b,c)=(x,y,z)$ by linear independence. 
A: So you're studying linear algebra. In that case lets rewrite $T(x,y,z)$ as in terms of a matrix times a vector:
$T(x,y,z)=
\begin{bmatrix}x-2y\\z\\x+y\end{bmatrix}=\begin{bmatrix}1&-2&0\\0&0&1\\1&1&0\end{bmatrix}\cdot\begin{bmatrix}x\\y\\z\end{bmatrix}
$
Now there is a number of ways to show that $T$ is bijective. The easiest by far is to show that the determinant of its matrix is different from $0$. If you haven't learned about determinants yet you could try to reduce the matrix with row operations to an identity matrix and that implies that it is bijective. 
If you have showed that the $T$ is injective, you can prove it is surjective this way:
Let $\bar{x_0}=(x_0,y_0,z_0)$ be an element of your codomain. Lets try to find the element in $\bar{x}=(x,y,z)$ in your domain, so that $T(\bar{x})=\bar{x_0}$. Now.. if $\bar{x}$ exist, it must forfill that:
$\begin{bmatrix}x-2y\\z\\x+y\end{bmatrix}=\begin{bmatrix}x_0\\y_0\\z_0\end{bmatrix}$
From this equation we see that:
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}\frac{x_0+2z_0}{3}\\z_0\\\frac{z_0-x_0}{3}\end{bmatrix}$
Now the question is if $\begin{bmatrix}\frac{x_0+2z_0}{3}\\z_0\\\frac{z_0-x_0}{3}\end{bmatrix}$ lies in your domain. If it does, then $T$ is surjective.
