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(I have very little background in category theory)

I learned that $-\otimes N$ and $N\otimes -$ preserves cokernel, that is given an exact sequence of $R$-modules $$A\xrightarrow{f} B \xrightarrow {g} C\rightarrow 0,$$ we have $$A\otimes N\xrightarrow{f\otimes id_N} B\otimes N \xrightarrow {g\otimes id_N} C\otimes N \rightarrow 0$$ and $$N\otimes A\xrightarrow{id_N \otimes f} N\otimes B \xrightarrow {id_N \otimes g} N\otimes C \rightarrow 0$$ are also exact.

And $\hom(-,N)$ and $\hom(N,-)$ preserve kernels. Again, given an exact sequence, $$0\rightarrow A\xrightarrow{f} B \xrightarrow {g} C,$$ we know $$0\rightarrow \hom(N,A ) \xrightarrow{ f \circ } \hom(N,B) \xrightarrow{ g \circ } \hom(N,C)$$ are exact. And if $$A\xrightarrow{f} B\xrightarrow {g} C \rightarrow 0$$ is exact, then

$$0\rightarrow \hom(C,N ) \xrightarrow{\circ g} \hom(B,N) \xrightarrow{\circ f} \hom(A,N)$$ is exact.

In the above statements, why do we need switch the orders of $A,B,C$ in the exact sequence for $\hom(-,N)$ and $\hom(N,-)$ but not necessarily for $-\otimes N$ and $N\otimes -$. Is it because of $M\otimes N \cong N\otimes M$ but $\hom(M,N)\not\cong\hom(N,M)$ in general?

I know tensor and $\hom$ should be closely related, so what caused this difference, at least for vector spaces, $V^*\otimes W \cong \hom(V,W)$.

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    $\text{Hom}(-, N)$ converts cokernels to kernels. – 2017-03-07

1 Answers 1

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This is because tensor product is a covariant functor in both variables, where as Hom is covariant in the second variable, and contravariant in the first variable. As to $V^*\otimes W$, it is contravariant in the first variable, since it results from the composition of the (contravariant) dual functor with the (covariant) tensor product.