I am currently taking a course in statistics, and in this course we are considering linear models $\mu = X\beta$ where $\mu \in L$ and $L = col(X)$ is a linear subspace of $\mathbb{R}^n$, $X$ is the design matrix and $\beta$ are the parameters. In this course, our lecturer has repeatedly emphasised that there is a one-to-one correspondence between $\mu$ and $\beta$ if and only if $X$ has full column rank.
However, I am simply not able to see how this can be the case. My reasoning is as follows. If there is a one-to-one correspondence between $\mu$ and $\beta$, then $X$ must be bijective, i.e. be both injective and surjective. Injectivity follows quite easily from the fact that $X$ has full column rank. According to my limited knowledge of linear algebra, however, a matrix is surjective if and only if it has full row rank, which, together with injectivity means that $X$ is square, but the design matrix is certainly not square in general, in fact it almost always has more rows than columns. I have a vague idea that surjectivity may somehow follow from the fact that $L = col(X)$ and $L$ is a linear subspace, but I can't see exactly how this avoids the problem of making the design matrix square.
Where is my thinking wrong and how can I prove that there is a one-to-one correspondence between $\mu$ and $\beta$?
Having looked into it some more, I have found an answer with the help of drhab's comments.
Injectivity follows from the fact that the columns of $X$ are linearly independent, so if $X\beta_1 = X\beta_2$, then $X(\beta_1 - \beta_2) = 0$, so $\beta_1 = \beta_2$, and $X$ is injective.
Surjectivity follows from the fact that $X$ maps to the linear subspace $L = col(X) = \{ \alpha_1 X_1 + \dots + \alpha_k X_k \; | \; (\alpha_1, \dots, \alpha_k) \in \mathbb{R}^k\}$, and we see that $X\beta = \beta_1 X_1 + \dots + \beta_k X_k$, so all of $L$ is hit by $X$, and $X$ is therefore surjective.