why are deterministic PDAs not closed under concatenation?
Pick the languages:

$L_1 = \{a^ib^jc^k | i \neq j \}$ and $L_2 = \{a^ib^jc^k | j \neq k\}$; both are DCFL and $L_3 = 0L_1 \cup L_2$ is DCFL, too.

$L_0 = 0^*$ is DCFL (regular)

But $L_{conc} = L_0 \cdot L_3 = 0^* L_3$ is not DCFL.

Proof: Suppose that $L_{conc}$ (which is the concatenation of two DCFLs) is DCFL.

If we intersect $L_{conc}$ with the regular language $0a^*b^*c^*$, we should get a DCFL language:

$L_{conc} \cap \{0a^*b^*c^*\}$ = $0L_1 \cup 0L_2$. suppose $0L_1 \cup 0L_2$ is a DCFL, so $L_1 \cup L_2$ should be a DCFL, too but:

$L_1 \cup L_2 = \overline{\overline{L_1} \cap \overline{L_2}} = \overline{\{a^ib^ic^i\}}$ which is not DCFL $\Rightarrow$ contradiction.