# The existence of "inverse" in monoidal category

In a monoidal category $$\mathcal{C}$$, Does any $$f\in \operatorname{Hom}_{\mathcal{C}}(X\otimes \mathbf{1},Y\otimes \mathbf{1})$$ can be expressed as $$f=g\otimes \operatorname{Id}_{\mathbf{1}}$$, where $$g\in\operatorname{Hom}_{\mathcal{C}}(X,Y)$$?

Short remark: I am reading a lecture note about tensor category by P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik. The link is attached here https://ocw.mit.edu/courses/18-769-topics-in-lie-theory-tensor-categories-spring-2009/pages/lecture-notes/. In their convention, a monoidal category is defined as a quintuple $$(\mathcal{C},\otimes, a, \mathbf{1}, \iota)$$ where $$\mathcal{C}$$ is a category, $$\otimes : \mathcal{C}\times \mathcal{C}\to \mathcal{C}$$ is a bifunctor.

$$a: (\bullet\otimes \bullet)\otimes \bullet \xrightarrow[]{\sim} \bullet\otimes(\bullet\otimes\bullet)$$ is natural isomorphism between two tri-functor $$\mathcal{C}\times\mathcal{C}\times\mathcal{C}\to \mathcal{C}$$. $$a_{X,Y,Z}: (X\otimes Y)\otimes Z \xrightarrow[]{\sim} X\otimes (Y\otimes Z),\quad X,Y,Z\in\mathcal{C}$$ called the associativity constraint (or associativity isomorphism). $$\mathbf{1}\in\mathcal{C}$$ is an object of $$\mathcal{C}$$, and $$\iota: \mathbf{1}\otimes \mathbf{1}\to \mathbf{1}$$ is an isomorphism, subject to the following two axioms.

1. The pentagon axiom.

2. The unit axiom. We can easily prove that $$X\to L_1(X):=\mathbf{1}\otimes X$$, $$X\in \mathcal{C}$$, Any $$f\in \operatorname{Hom}_{\mathcal{C}}(X,Y)$$ is mapped to $$L_1(f):= \operatorname{Id}_{\mathbf{1}}\otimes f$$, $$L_1$$ defines a functor $$\mathcal{C}\to \mathcal{C}$$. The functor $$R_1$$ is defined similarly. The functors $$L_{\mathbf{1}}$$ and $$R_{\mathbf{1}}$$ of left and right multiplication by $$\mathbf{1}$$ are equivalences $$\mathcal{C}\to \mathcal{C}$$.

The unit axiom provides the following commutative diagram (natural isomorphism $$\eta: (R_1^{-1}\circ R_1)\xrightarrow{\sim} \operatorname{Id}_{\mathcal{C}}$$)

$$\require{AMScd} \begin{CD} R_1^{-1}(X\otimes \mathbf{1}) @>{\eta_X}>> X\\ @V{R_1^{-1}(g\otimes \operatorname{Id}_{\mathbf{1}})}VV @VV{g}V \\ R_1^{-1}(Y\otimes \mathbf{1}) @>{\eta_Y}>> Y \end{CD}$$

Therefore, if such $$g$$ exists, it can be written as $$g=\eta_Y\circ R_1^{-1}(f)\circ \eta_X^{-1}$$.

The answer is yes. We can prove that $$f=(\eta_Y\circ R_1^{-1}(f)\circ\eta_X^{-1})\otimes\operatorname{Id}_{\mathbf{1}}\quad \forall f\in\operatorname{Hom}_{\mathcal{C}}(X,Y)$$

Let $$g=\eta_Y\circ R_1^{-1}(f)\circ\eta_X^{-1}$$ in the commutative diagram present in the question, we will get $$R_1^{-1}(g\otimes \operatorname{Id}_{\mathbf{1}})=R_1^{-1}(f)$$. As mentioned in the lecture note attached in the question, there exists a natural isomorphism $$r: R_1\xrightarrow{\sim}\operatorname{Id}_{\mathcal{C}}$$, combined with the natural isomorphism (category equivalence) $$\eta^\prime: (R_1\circ R_1^{-1})\xrightarrow{\sim}\operatorname{Id}_{\mathcal{C}}$$, we can establish a natural isomorphism $$\lambda: R_1^{-1}\xrightarrow{\sim}\operatorname{Id}_{\mathcal{C}}$$. This natural isomorphism tells that if $$R_1^{-1}(g\otimes \operatorname{Id}_{\mathbf{1}})=R_1^{-1}(f)$$ then $$g\otimes\operatorname{Id}_{\mathbf{1}}=f$$.