CFG ambiguity is undecidable

Choose fresh terminal symbols ${\iota }_1,\dots ,{\iota }_n$ disjoint from the PCP alphabet, and build one grammar with start symbol $S$ and two branches: $S\to T\mid B,$ $T\to u_{i}T{\iota }_i\mid u_i{\iota }_i,B\to v_{i}B{\iota }_i\mid v_i{\iota }_i(1\le i\le n)\mathrm{.}$ A leftmost derivation from $T$ yields exactly the words $u_{i_1}\cdots u_{i_k}{\iota }_{i_k}\cdots {\iota }_{i_1}$, and a derivation from $B$ yields the analogous words with the $v_i$. The fresh reversed suffix uniquely determines the entire index sequence, so each branch is unambiguous.

If $i_1,\dots ,i_k$ is a PCP match, then the word $u_{i_1}\cdots u_{i_k}{\iota }_{i_k}\cdots {\iota }_{i_1}=v_{i_1}\cdots v_{i_k}{\iota }_{i_k}\cdots {\iota }_{i_1}$ has one leftmost derivation beginning with $S\Rightarrow T$ and another beginning with $S\Rightarrow B$. Hence the grammar is ambiguous by [L1].

Conversely, if the grammar is ambiguous, [L1] gives a word with two distinct leftmost derivations. Step 1.1 proves that neither the $T$ branch nor the $B$ branch has two derivations of one word, so one derivation begins with $S\Rightarrow T$ and the other with $S\Rightarrow B$. The common terminal word has the form $u_{i_1}\cdots u_{i_k}{\iota }_{i_k}\cdots {\iota }_{i_1}=v_{j_1}\cdots v_{j_m}{\iota }_{j_m}\cdots {\iota }_{j_1}\mathrm{.}$ Freshness and unique parsing of the reversed suffix force $k=m$ and $i_r=j_r$ for every $r$, so $i_1,\dots ,i_k$ is a PCP match.

Steps 2.1 and 2.2 show that the constructed grammar is ambiguous if and only if the original PCP instance has a match. If CFG ambiguity were decidable, then [L4] and this equivalence would make PCP decidable, contradicting [L3].

Therefore CFG ambiguity is undecidable.