Prove $\textbf{F}^{\{1,2\}}$ and $\textbf{F}^{2}$ are isomorphic

Question

I know this is a rather trivial isomorphism, but I am going to give a proof a shot.Is this completely correct?

Define a linear map $T:\textbf{F}^{\{1,2\}} \rightarrow \textbf{F}^2$ by $T(f)=(f(1),f(2))$

to prove injectivity:

Suppose $T(f)=(0,0) \implies f=0$ where $f$ is the zero function in $\textbf{F}^{\{1,2\}}$

So null$T=\{0\}$ and $T$ is injective.

to prove surjectivity:

Suppose $(x,y) \in \textbf{F}^2$

Let $f:\{1,2\} \rightarrow \textbf{F}$ such that $f(1)=x,f(2)=y$

Thus every $(x,y) \in \textbf{F}^2$ is in the range of $T$, proving surjectivity.

TO finish off the proof prove $T$ is linear.

Let $f,g \in \textbf{F}^{\{1,2\}},c,d \in \textbf{F} $

$T(cf+dg)=((cf+dg)(1),(cf+dg)(2))=(cf(1),cf(2))+(dg(1),dg(2))=c(f(1),f(2))+d(g(1),g(2))$

$=cT(f)+dT(g)$

Answer 1

Your proof looks fine to me! Well done!