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There's even a more direct way than the one described by Adam. Note that every pure Choi state corresponds to a superoperator acting by conjugation. Since $$ \mathbb{I} \otimes \mathbb{I} |\phi^+\rangle = |\phi^+\rangle, \qquad Z \otimes \mathbb{I} |\phi^+\rangle = |\phi^-\rangle, $$ the inverse of $c_\lambda$ under the Choi-Jamiołkowski isomorphism is simply $$ T_\lambda = \lambda\, \mathrm{id} + (1-\lambda)\, Z \cdot Z^\dagger. $$ This gives you the Kraus operators $$ K_1 := \sqrt{\lambda}\,\mathbb{I}, \qquad K_2:= \sqrt{1-\lambda}\, Z. $$

Remark: In general, you can find the Kraus operators by "reshaping" the pure Choi states, since $$ \mathcal{J}(K\cdot K^\dagger) = \mathrm{vec}(K)\mathrm{vec}(K)^\dagger, $$ where the vectorisation isomorphism $\mathrm{vec}:\, L(\mathcal H)\simeq \mathcal H\otimes \mathcal H^* \rightarrow \mathcal H\otimes \mathcal H$ acts as $$ \mathrm{vec}(|x\rangle\langle y|) = |x\rangle \otimes | y\rangle $$ i.e. it applies the Riesz isomorphism to the second factor.

There's even a more direct way than the one described by Adam. Note that every pure Choi state corresponds to a superoperator acting by conjugation. Since $$ \mathbb{I} \otimes \mathbb{I} |\phi^+\rangle = |\phi^+\rangle, \qquad Z \otimes \mathbb{I} |\phi^+\rangle = |\phi^-\rangle, $$ the inverse of $c_\lambda$ under the Choi-Jamiołkowski isomorphism is simply $$ T_\lambda = \lambda\, \mathrm{id} + (1-\lambda)\, Z \cdot Z^\dagger. $$ This gives you the Kraus operators $$ K_1 := \sqrt{\lambda}\,\mathbb{I}, \qquad K_2:= \sqrt{1-\lambda}\, Z. $$

Remark: In general, you can find the Kraus operators by "reshaping" the pure Choi states, since $$ \mathcal{J}(K\cdot K^\dagger) = d^{-1}\, \mathrm{vec}(K)\mathrm{vec}(K)^\dagger, $$ where the vectorisation isomorphism $\mathrm{vec}:\, L(\mathcal H)\simeq \mathcal H\otimes \mathcal H^* \rightarrow \mathcal H\otimes \mathcal H$ acts as $$ \mathrm{vec}(|x\rangle\langle y|) = |x\rangle \otimes | y\rangle $$ i.e. it applies the Riesz isomorphism to the second factor.

Edit: Sorry, I used two different conventions for the Choi-Jamiołkowski isomorphism, one where the image of a CPTP map has trace one and the other where it has trace $d$ (=dimension). Fixed the second part since OP uses first convention.

There's even a more direct way than the one described by Adam. Note that every pure Choi state corresponds to a superoperator acting by conjugation. Since $$ \mathbb{I} \otimes \mathbb{I} |\phi^+\rangle = |\phi^+\rangle, \qquad Z \otimes \mathbb{I} |\phi^+\rangle = |\phi^-\rangle, $$ the inverse of $c_\lambda$ under the Choi-Jamiołkowski isomorphism is simply $$ T_\lambda = \lambda\, \mathrm{id} + (1-\lambda)\, Z \cdot Z^\dagger. $$ This gives you the Kraus operators $$ K_1 := \sqrt{\lambda}\,\mathbb{I}, \qquad K_2:= \sqrt{1-\lambda}\, Z. $$

Remark: In general, you can find the Kraus operators by "reshaping" the pure Choi states, since $$ \mathcal{J}(K\cdot K^\dagger) = \mathrm{vec}(K)\mathrm{vec}(K)^\dagger, $$ where the vectorisation isomorphism $\mathrm{vec}:\, L(\mathcal H)\simeq \mathcal H\otimes \mathcal H^* \rightarrow \mathcal H\otimes \mathcal H$ acts as $$ \mathrm{vec}(|x\rangle\langle y|) = |x\rangle \otimes | y\rangle. $$ I.e. it applies the Riesz isomorphism to the second factor.

There's even a more direct way than the one described by Adam. Note that every pure Choi state corresponds to a superoperator acting by conjugation. Since $$ \mathbb{I} \otimes \mathbb{I} |\phi^+\rangle = |\phi^+\rangle, \qquad Z \otimes \mathbb{I} |\phi^+\rangle = |\phi^-\rangle, $$ the inverse of $c_\lambda$ under the Choi-Jamiołkowski isomorphism is simply $$ T_\lambda = \lambda\, \mathrm{id} + (1-\lambda)\, Z \cdot Z^\dagger. $$ This gives you the Kraus operators $$ K_1 := \sqrt{\lambda}\,\mathbb{I}, \qquad K_2:= \sqrt{1-\lambda}\, Z. $$

Remark: In general, you can find the Kraus operators by "reshaping" the pure Choi states, since $$ \mathcal{J}(K\cdot K^\dagger) = \mathrm{vec}(K)\mathrm{vec}(K)^\dagger, $$ where the vectorisation isomorphism $\mathrm{vec}:\, L(\mathcal H)\simeq \mathcal H\otimes \mathcal H^* \rightarrow \mathcal H\otimes \mathcal H$ acts as $$ \mathrm{vec}(|x\rangle\langle y|) = |x\rangle \otimes | y\rangle $$ i.e. it applies the Riesz isomorphism to the second factor.

There's even a more direct way than the one described by Adam. Note that every pure Choi state corresponds to a superoperator acting by conjugation. Since $$ \mathbb{I} \otimes \mathbb{I} |\phi^+\rangle = |\phi^+\rangle, \qquad Z \otimes \mathbb{I} |\phi^+\rangle = |\phi^-\rangle, $$ the inverse of $c_\lambda$ under the Choi-Jamiołkowski isomorphism is simply $$ T_\lambda = \lambda\, \mathrm{id} + (1-\lambda)\, Z \cdot Z^\dagger. $$ This gives you the Kraus operators $$ K_1 := \sqrt{\lambda}\,\mathbb{I}, \qquad K_2:= \sqrt{1-\lambda}\, Z. $$

There's even a more direct way than the one described by Adam. Note that every pure Choi state corresponds to a superoperator acting by conjugation. Since $$ \mathbb{I} \otimes \mathbb{I} |\phi^+\rangle = |\phi^+\rangle, \qquad Z \otimes \mathbb{I} |\phi^+\rangle = |\phi^-\rangle, $$ the inverse of $c_\lambda$ under the Choi-Jamiołkowski isomorphism is simply $$ T_\lambda = \lambda\, \mathrm{id} + (1-\lambda)\, Z \cdot Z^\dagger. $$ This gives you the Kraus operators $$ K_1 := \sqrt{\lambda}\,\mathbb{I}, \qquad K_2:= \sqrt{1-\lambda}\, Z. $$

Remark: In general, you can find the Kraus operators by "reshaping" the pure Choi states, since $$ \mathcal{J}(K\cdot K^\dagger) = \mathrm{vec}(K)\mathrm{vec}(K)^\dagger, $$ where the vectorisation isomorphism $\mathrm{vec}:\, L(\mathcal H)\simeq \mathcal H\otimes \mathcal H^* \rightarrow \mathcal H\otimes \mathcal H$ acts as $$ \mathrm{vec}(|x\rangle\langle y|) = |x\rangle \otimes | y\rangle. $$ I.e. it applies the Riesz isomorphism to the second factor.