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They need a pre-shared entanglement resource that permits deterministic production of a Bell pair. In particular, one shared Bell pair and two classical bits suffice for exact quantum teleportation. This condition is stronger than merely having some entanglement.

Let Alice's input be $|\psi {⟩}_C=\alpha |0⟩+\beta |1⟩$, and let her system $A$ and Bob's system $B$ share $|{\Phi }^{+}{⟩}_{AB}=(|00⟩+|11⟩)/\sqrt{2}$. Use the Bell states $|{\Phi }^{\pm }⟩=(|00⟩\pm |11⟩)/\sqrt{2}$ and $|{\Psi }^{\pm }⟩=(|01⟩\pm |10⟩)/\sqrt{2}$. Direct expansion givesAlice performs a projective measurement in this Bell basis and sends its two-bit outcome. Each outcome has probability $1/4$. Bob applies respectively $I,Z,X,ZX$, recovering $|\psi ⟩$ exactly, without either party knowing $\alpha ,\beta$. Before he receives the outcome, his averaged state is $I/2$, so the protocol does not communicate without the classical message. Alice's measurement consumes the input and the shared entanglement.

For necessity, exact transmission of every pure qubit means the induced quantum channel is the identity: pure-state projectors span the operator space, and a physical channel is linear. Alice can therefore prepare a Bell pair locally and transmit one half by the proposed protocol, leaving a Bell pair shared with Bob. The protocol would itself be a deterministic local operations and classical communication conversion of their resource into a Bell pair. This proves the exact teleportation resource criterion. With no shared entanglement, it is impossible: local operations and classical communication preserve separable quantum states, whereas the resulting Bell pair would be entangled.