how to prove the following is a bijection?
One option could be adding more parameters to $\chi$ so to make both $\Gamma$ and a fresh name source $\phi \in {\sf Names}^\infty$ explicit:

$$
\begin{array}{l}
\chi(\Gamma, x, X;\phi) = X \mbox{ if } x \not\in\Gamma
\\
\chi(\Gamma, x, \phi) = X \mbox{ if } x:X \in\Gamma
\\
\chi(\Gamma, \lambda x. t, X;\phi) = abs(X, \chi(\Gamma;x:X, t, \phi))
\\
\chi(\Gamma, t_1 t_2, \phi) = app(\chi(\Gamma, t_1, \phi^1),\chi(\Gamma, t_2, \phi^2))
\\
\qquad \mbox{where } \phi = \phi^1_0 \phi^2_0 \phi^1_1 \phi^2_1 \phi^1_2 \phi^2_2 \ldots
\end{array}
$$

You can then state a form of injectivity as
$$
\chi(\Gamma, t_1, \phi) = \chi(\Gamma, t_2, \phi) \implies t_1 = t_2
$$
and surjectivity as
$$
\forall T.\ \exists \Gamma, t, \phi.\ \chi(\Gamma, t, \phi) = T
$$

Note that some stronger forms of injectivity do not hold. E.g.
$$
\chi(\Gamma_1, t_1, \phi) = \chi(\Gamma_2, t_2, \phi) \implies \Gamma_1 = \Gamma_2 \land t_1 = t_2
$$
is false since 
$$
\chi(x:X,x, \phi) = \chi(y:X,y, \phi) \land x:X \neq y:X
$$
Similarly, this does not hold:
$$
\chi(\Gamma, t, \phi_1) = \chi(\Gamma, t, \phi_2) \implies \phi_1 = \phi_2
$$
since only a finite portion on $\phi$ is needed by $\chi$.