In a closed symmetric monoidal category with $[I,X] \cong X$ for all $X$ is it true that having monomorphisms $m :A \rightarrow B$ and $m: B \rightarrow A$ is enough to imply $A \cong B$ ?
I tried to figure this out, and expected to end up using the Yoneda Embedding, problem is, my proof seems to hold for all categories, which is clearly wrong. Here is my attempt!
Given $m: A \rightarrow B$, using the covariant hom functor $Hom(X,-):C \rightarrow SET$ there is a morphism $Hom(X,m): Hom(X,A) \rightarrow Hom(X,B)$ Defined by $Hom(X,m)(h) \equiv m \circ h$.
Since $m$ is a monomorphism $Hom(X,m)(h_1) = Hom(X,m)(h_2) \iff m \circ h_1 = m \circ h_2 \iff h_1 = h_2$. Since there are both ways monomorphisms, there are both ways injections, so there is a bijection between $Hom(X,m)(h_1)$ and $Hom(X,m)(h_2)$. Any injection between the two must then be a bijection.
Now $M_i \equiv Hom(X_{i},m)$ form the components of a natural transformation $M: Hom(-,A) \rightarrow Hom(-,B)$, since for any $h \in Hom(X_1,A)$ and $f: X_2 \rightarrow X_1$ then $Hom(f,B)(Hom(X_1,m)(h)) = m \circ h \circ f = Hom(X_2,m)(Hom(f,A)(h))$.
So $M$ is a natural transformation with every component a bijection, a natural isomorphism. So then $Hom(-,A) \cong Hom(-,B)$ and by Yoneda $A \cong B$.
