The existence of "inverse" in monoidal category

Question

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 $g=\eta_Y\circ R_1^{-1}(f)\circ \eta_X^{-1}$.

Answer 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 $g\otimes\operatorname{Id}_{\mathbf{1}}=f$.