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For a monad $(T,\mu,\eta)$ if $T(A) = T(B)$, does this imply that $\mu_A = \mu_B$? I want to know because in the bijection between Kleisli categoriestriples and monads, given a monad, we define $f^* := T(f) ; \mu_B$ if $f : A\to T(B)$ (c.f. WikipediaProp 1.6), but this needs to be well defined even when $T$ is not an embedding.

Clarification (more formal): Given a monad $(T,\mu,\eta)$, I need to define ${}^*$ on the class $\{f \;|\; \exists A,B \in |\mathbf{C}| . f : A \to T B \}$. If $T$ were an embedding then I could just take $B := T^{-1}(\mathrm{cod}(f))$. What do I do though when $T$ is not an embedding? All I know about the codomain of an element of this class is that it is in the image of $T$. There may be many $B$'s and the $\mu_B$'s may be different!

For a monad $(T,\mu,\eta)$ if $T(A) = T(B)$, does this imply that $\mu_A = \mu_B$? I want to know because in the bijection between Kleisli categories and monads, given a monad, we define $f^* := T(f) ; \mu_B$ if $f : A\to T(B)$ (c.f. Wikipedia), but this needs to be well defined even when $T$ is not an embedding.

Clarification (more formal): Given a monad $(T,\mu,\eta)$, I need to define ${}^*$ on the class $\{f \;|\; \exists A,B \in |\mathbf{C}| . f : A \to T B \}$. If $T$ were an embedding then I could just take $B := T^{-1}(\mathrm{cod}(f))$. What do I do though when $T$ is not an embedding? All I know about the codomain of an element of this class is that it is in the image of $T$. There may be many $B$'s and the $\mu_B$'s may be different!

For a monad $(T,\mu,\eta)$ if $T(A) = T(B)$, does this imply that $\mu_A = \mu_B$? I want to know because in the bijection between Kleisli triples and monads, given a monad, we define $f^* := T(f) ; \mu_B$ if $f : A\to T(B)$ (c.f. Prop 1.6), but this needs to be well defined even when $T$ is not an embedding.

Clarification (more formal): Given a monad $(T,\mu,\eta)$, I need to define ${}^*$ on the class $\{f \;|\; \exists A,B \in |\mathbf{C}| . f : A \to T B \}$. If $T$ were an embedding then I could just take $B := T^{-1}(\mathrm{cod}(f))$. What do I do though when $T$ is not an embedding? All I know about the codomain of an element of this class is that it is in the image of $T$. There may be many $B$'s and the $\mu_B$'s may be different!

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Neel Krishnaswami
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For a monad $(T,\mu,\eta)$ if $T(A) = T(B)$, does this imply that $\mu_A = \mu_B$? I want to know because in the bijection between Kleisli categories and monads, given a monad, we define $f^* := T(f) ; \mu_B$ if $f : A\to T(B)$ (c.f. Wikipedia), but this needs to be well defined even when $T$ is not an embedding.

Clarification (more formal): Given a monad $(T,\mu,\eta)$, I need to define ${}^*$ on the class $\{f|\exists A,B \in |\mathbf{C}| . f : A \to T B \}$$\{f \;|\; \exists A,B \in |\mathbf{C}| . f : A \to T B \}$. If $T$ were an embedding then I could just take $B := T^{-1}(\mathrm{cod}(f))$. What do I do though when $T$ is not an embedding? All I know about the codomain of an element of this class is that it is in the image of $T$. There may be many $B$'s and the $\mu_B$'s may be different!

For a monad $(T,\mu,\eta)$ if $T(A) = T(B)$, does this imply that $\mu_A = \mu_B$? I want to know because in the bijection between Kleisli categories and monads, given a monad, we define $f^* := T(f) ; \mu_B$ if $f : A\to T(B)$ (c.f. Wikipedia), but this needs to be well defined even when $T$ is not an embedding.

Clarification (more formal): Given a monad $(T,\mu,\eta)$, I need to define ${}^*$ on the class $\{f|\exists A,B \in |\mathbf{C}| . f : A \to T B \}$. If $T$ were an embedding then I could just take $B := T^{-1}(\mathrm{cod}(f))$. What do I do though when $T$ is not an embedding? All I know about the codomain of an element of this class is that it is in the image of $T$. There may be many $B$'s and the $\mu_B$'s may be different!

For a monad $(T,\mu,\eta)$ if $T(A) = T(B)$, does this imply that $\mu_A = \mu_B$? I want to know because in the bijection between Kleisli categories and monads, given a monad, we define $f^* := T(f) ; \mu_B$ if $f : A\to T(B)$ (c.f. Wikipedia), but this needs to be well defined even when $T$ is not an embedding.

Clarification (more formal): Given a monad $(T,\mu,\eta)$, I need to define ${}^*$ on the class $\{f \;|\; \exists A,B \in |\mathbf{C}| . f : A \to T B \}$. If $T$ were an embedding then I could just take $B := T^{-1}(\mathrm{cod}(f))$. What do I do though when $T$ is not an embedding? All I know about the codomain of an element of this class is that it is in the image of $T$. There may be many $B$'s and the $\mu_B$'s may be different!

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For a monad $(T,\mu,\eta)$ if $T(A) = T(B)$, does this imply that $\mu_A = \mu_B$? I want to know because in the bijection between Kleisli categories and monads, given a monad, we define $f^* := T(f) ; \mu_B$ if $f : A\to T(B)$ (c.f. Wikipedia), but this needs to be well defined even when $T$ is not an embedding.

Clarification (more formal): Given a monad $(T,\mu,\eta)$, I need to define ${}^*$ on the class $\{f|\exists A,B \in |\mathbf{C}| . f : A \to T B \}$. If $T$ were an embedding then I could just take $B := T^{-1}(\mathrm{cod}(f))$. What do I do though when $T$ is not an embedding? All I know about the codomain of an element of this class is that it is in the image of $T$. There may be many $B$'s and the $\mu_B$'s may be different!

For a monad $(T,\mu,\eta)$ if $T(A) = T(B)$, does this imply that $\mu_A = \mu_B$? I want to know because in the bijection between Kleisli categories and monads, given a monad, we define $f^* := T(f) ; \mu_B$ if $f : A\to T(B)$, but this needs to be well defined even when $T$ is not an embedding.

Clarification (more formal): Given a monad $(T,\mu,\eta)$, I need to define ${}^*$ on the class $\{f|\exists A,B \in |\mathbf{C}| . f : A \to T B \}$. If $T$ were an embedding then I could just take $B := T^{-1}(\mathrm{cod}(f))$. What do I do though when $T$ is not an embedding? All I know about the codomain of an element of this class is that it is in the image of $T$. There may be many $B$'s and the $\mu_B$'s may be different!

For a monad $(T,\mu,\eta)$ if $T(A) = T(B)$, does this imply that $\mu_A = \mu_B$? I want to know because in the bijection between Kleisli categories and monads, given a monad, we define $f^* := T(f) ; \mu_B$ if $f : A\to T(B)$ (c.f. Wikipedia), but this needs to be well defined even when $T$ is not an embedding.

Clarification (more formal): Given a monad $(T,\mu,\eta)$, I need to define ${}^*$ on the class $\{f|\exists A,B \in |\mathbf{C}| . f : A \to T B \}$. If $T$ were an embedding then I could just take $B := T^{-1}(\mathrm{cod}(f))$. What do I do though when $T$ is not an embedding? All I know about the codomain of an element of this class is that it is in the image of $T$. There may be many $B$'s and the $\mu_B$'s may be different!

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