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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 $A \cong B$.

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    $\begingroup$ If you have two monomorphisms X -> Y and Y -> X in any category, then Hom(X, Z) = Hom(Y, Z). The issue is that this cannot necessarily be made natural in Z, so X need not be isomorphic to Y. I haven't read what you wrote, but I'm guessing some overlooking of naturality is where the problem shows up. $\endgroup$
    – skd
    Apr 3, 2020 at 0:43
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    $\begingroup$ It's certainly very far from true. Just take for example the category of compactly generated spaces, and consider the spaces $[0, 1]$ and $(0, 1)$. For another simple example, take the category of abelian groups, and the coproduct of $\mathbb{Z}$ and countably many copies of $\mathbb{Q}$. $\endgroup$
    – Todd Trimble ♦
    Apr 3, 2020 at 0:54
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    $\begingroup$ Your proof first shows, using Cantor-Bernstein in Set, that there is a bijection between hom(X,A) and hom(X,B). However, after that you claim that "any injection between the two must then be a bijection" - this is false if the hom-sets are infinite. $\endgroup$
    – Martti Karvonen
    Apr 3, 2020 at 1:03

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