Let's recall a simple, elementary, and general fact that hasn't been explicitly mentioned: a dual Banach space is always a splitting subspace in the isometric embedding into its double dual.

Let $i_X:X\to X^{**}$ denote the natural isometric embedding of $X$ in $X^{**}$. If we dualize, we have a transpose operator, $i_X^*:X^{***}\to X^*$ (that we may identify as the restriction map, which takes a linear form on $X^{**}$ to its restriction on $X$ as a subspace of $X^{**}$). On the other hand we also have the isometric embedding $i_{X^*}:X^*\to X^{***}$. It is a straightforward (though a bit formal) computation checking that $i_{X}^*$ is left-inverse to $i_{X^*}$, that is $i_{X}^*i_{X^*}=1_{X^*}.$ As a consequence of this, $P:=i_{X^*}i_{X}^*$ is a linear projector with $\operatorname{ker}P=\operatorname{ker}i_X^*=X^\perp$ corresponding to the splitting $X^{***}=X^*\oplus X^\perp$.

$$*$$

Checking the identity $i_{X}^*i_{X^*}=1_{X^*}.$ This means $i_{X}^*i_{X^*}f=f$ for all $f\in X^*$, which also means $\langle i_{X}^*i_{X^*}f, x\rangle=\langle f, x\rangle$ for all $x\in X$ and $f\in X^*$. Indeed

$$\langle i_{X}^*i_{X^*}f, x\rangle_{X^*\times X}=\langle i_{X^*}f, i_{X} x\rangle_{X^{***}\times X^{**}}=\langle i_{X} x,f\rangle_{X^{**}\times X^*}=\langle f, x\rangle_{X^*\times X},$$ by the definition of the transpose operator $i_{X}^*$, respectively by the definition of the embeddings $i_{X^*}$ and $i_{X}$.

Let's recall a simple, elementary, and general fact that hasn't been explicitly mentioned: a dual Banach space is always a splitting subspace in the isometric embedding into its double dual. So $\ell_1$ is a particular case, for $X=c_0$.

Let $i_X:X\to X^{**}$ denote the natural isometric embedding of $X$ in $X^{**}$. If we dualize, we have a transpose operator, $i_X^*:X^{***}\to X^*$ (that we may identify as the restriction map, which takes a linear form on $X^{**}$ to its restriction on $X$ as a subspace of $X^{**}$). On the other hand we also have the isometric embedding $i_{X^*}:X^*\to X^{***}$. It is a straightforward (though a bit formal) computation checking that $i_{X}^*$ is left-inverse to $i_{X^*}$, that is $i_{X}^*i_{X^*}=1_{X^*}.$ As a consequence of this, $P:=i_{X^*}i_{X}^*$ is a linear projector with $\operatorname{ker}P=\operatorname{ker}i_X^*=X^\perp$ corresponding to the splitting $X^{***}=X^*\oplus X^\perp$.

$$*$$

Checking the identity $i_{X}^*i_{X^*}=1_{X^*}.$ This means $i_{X}^*i_{X^*}f=f$ for all $f\in X^*$, which also means $\langle i_{X}^*i_{X^*}f, x\rangle=\langle f, x\rangle$ for all $x\in X$ and $f\in X^*$. Indeed

$$\langle i_{X}^*i_{X^*}f, x\rangle_{X^*\times X}=\langle i_{X^*}f, i_{X} x\rangle_{X^{***}\times X^{**}}=\langle i_{X} x,f\rangle_{X^{**}\times X^*}=\langle f, x\rangle_{X^*\times X},$$ by the definition of the transpose operator $i_{X}^*$, respectively by the definition of the embeddings $i_{X^*}$ and $i_{X}$.

Let's recall the simple, elementary, and general fact that hasn't been explicitly mentioned: a dual Banach space is always a splitting subspace in the isometric embedding into its double dual.

Let $i_X:X\to X^{**}$ denote the natural isometric embedding of $X$ in $X^{**}$. If we dualize, we have a transpose operator, $i_X^*:X^{***}\to X^*$ (that we may identify as the restriction map, which takes a linear form on $X^{**}$ to its restriction on $X$ as a subspace of $X^{**}$). On the other hand we also have the isometric embedding $i_{X^*}:X^*\to X^{***}$. It is a straightforward (though a bit formal) computation checking that $i_{X}^*$ is left-inverse to $i_{X^*}$, that is $i_{X}^*i_{X^*}=1_{X^*}.$ As a consequence of this, $P:=i_{X^*}i_{X}^*$ is a linear projector with $\operatorname{ker}P=\operatorname{ker}i_X^*=X^\perp$ corresponding to the splitting $X^{***}=X^*\oplus X^\perp$.

$$*$$

Checking the identity $i_{X}^*i_{X^*}=1_{X^*}.$ Talking about an equality of maps, this means $i_{X}^*i_{X^*}f=f$ for all $f\in X^*$, which also means $\langle i_{X}^*i_{X^*}f, x\rangle=\langle f, x\rangle$ for all $x\in X$ and $f\in X^*$. Indeed

$$\langle i_{X}^*i_{X^*}f, x\rangle_{X^*\times X}=\langle i_{X^*}f, i_{X} x\rangle_{X^{***}\times X^{**}}=\langle i_{X} x,f\rangle_{X^{**}\times X^*}=\langle f, x\rangle_{X^*\times X},$$ by the definition of the transpose operator $i_{X}^*$, respectively by the definition of the embeddings $i_{X^*}$ and $i_{X}$.

Let's recall a simple, elementary, and general fact that hasn't been explicitly mentioned: a dual Banach space is always a splitting subspace in the isometric embedding into its double dual.

Let $i_X:X\to X^{**}$ denote the natural isometric embedding of $X$ in $X^{**}$. If we dualize, we have a transpose operator, $i_X^*:X^{***}\to X^*$ (that we may identify as the restriction map, which takes a linear form on $X^{**}$ to its restriction on $X$ as a subspace of $X^{**}$). On the other hand we also have the isometric embedding $i_{X^*}:X^*\to X^{***}$. It is a straightforward (though a bit formal) computation checking that $i_{X}^*$ is left-inverse to $i_{X^*}$, that is $i_{X}^*i_{X^*}=1_{X^*}.$ As a consequence of this, $P:=i_{X^*}i_{X}^*$ is a linear projector with $\operatorname{ker}P=\operatorname{ker}i_X^*=X^\perp$ corresponding to the splitting $X^{***}=X^*\oplus X^\perp$.

$$*$$

Checking the identity $i_{X}^*i_{X^*}=1_{X^*}.$ This means $i_{X}^*i_{X^*}f=f$ for all $f\in X^*$, which also means $\langle i_{X}^*i_{X^*}f, x\rangle=\langle f, x\rangle$ for all $x\in X$ and $f\in X^*$. Indeed

$$\langle i_{X}^*i_{X^*}f, x\rangle_{X^*\times X}=\langle i_{X^*}f, i_{X} x\rangle_{X^{***}\times X^{**}}=\langle i_{X} x,f\rangle_{X^{**}\times X^*}=\langle f, x\rangle_{X^*\times X},$$ by the definition of the transpose operator $i_{X}^*$, respectively by the definition of the embeddings $i_{X^*}$ and $i_{X}$.