Outline:

Given any function $f$ purporting to violate the theorem by being surjective (or cofinal) on $\lambda$ , we show that there are fewer than $\kappa$ possible values of $f(\alpha)$ , and therefore only $\max(\alpha,\kappa)$ possible elements in the entire range of $f$ , so $f$ is not surjective (or cofinal).

Details:

Suppose $\lambda>\kappa$ is a cardinal of $\mathfrak{M}$ that is not a cardinal in $\mathfrak{M}[G]$ .

There is some function $f\in\mathfrak{M}[G]$ and some cardinal $\alpha<\lambda$ such that $f:\alpha\rightarrow\lambda$ is surjective. This has a name, $\hat{f}$ . For each $\beta<\alpha$ , consider $$F_\beta=\{\gamma<\lambda\mid p\Vdash \hat{f}(\beta)=\gamma\}\text{ for some }p\in P$$

$|F_\beta|<\kappa$ , since any two $p\in P$ which force different values for $\hat{f}(\beta)$ are incompatible and $P$ has no sets of incompatible elements of size $\kappa$ .

Notice that $F_\beta$ is definable in $\mathfrak{M}$ . Then the range of $f$ must be contained in $F=\bigcup_{i<\alpha} F_i$ . But $|F|\leq\alpha\cdot\kappa=\max(\alpha,\kappa)<\lambda$ . So $f$ cannot possibly be surjective, and therefore $\lambda$ is not collapsed.

Now suppose that for some $\alpha\geq\lambda>\kappa$ , $\operatorname{cf}(\alpha)=\lambda$ in $\mathfrak{M}$ and for some $\eta<\lambda$ there is a cofinal function $f:\eta\rightarrow\alpha$ .

We can construct $F_\beta$ as above, and again the range of $f$ is contained in $F=\bigcup_{i<\eta} F_i$ . But then $|\operatorname{range}(f)|\leq|F|\leq\eta\cdot\kappa<\lambda$ . So there is some $\gamma<\alpha$ such that $f(\beta)<\gamma$ for any $\beta<\eta$ , and therefore $f$ is not cofinal in $\alpha$ .