# Evaluation map of internal hom of chain complexes

In the category of chain complexes over some abelian category, besides the ordinary external $\hom(C^*, D^*)$ which contains chain maps, there is the internal hom which is the chain complex

$\hom^i(C^*, D^*) = \prod_j \hom(C^j, D^{i+j})$

with the appropriate differential. Now, an internal hom admits the evaluation map

$\hom^*(C^*, D^*)\otimes C^*\rightarrow D^*$.

What is this map explicitly? My problem is the following: Let $C^*=D^*=(C^0\stackrel{d}{\rightarrow}C^1)$. Now $\hom^*(C^*, C^*) = \{1\}^0\rightarrow \{d\}^1$ and

$\hom^*(C^*, C^*)\otimes C^* = \{1\}\otimes C^0\rightarrow \{d\}\otimes C^0\oplus \{1\}\otimes C^1 \rightarrow \{d\}\otimes C^1$,

starting in degree 0 at the left. But what does the evaluation map do in degree 2, since $d$ is defined on $C^0$?

I suppose there is a profound misunderstanding how I should regard $\hom^*$, so maybe someone could clarify this?

So, I believe you work in the category of $R$-modules for some commutative ring, or any abelian category with a monoidal structure and an internal hom. Otherwise, there are a priori no internal hom in the category of chain complexes.

If $C^*=D^*$ is as you described, the internal hom you wrote is not correct. It is rather : $$\operatorname{hom}^*(C^*,C^*)=\operatorname{hom}(C^0,C^0)\times \operatorname{hom}(C^1,C^1)\rightarrow \operatorname{hom}(C^0,C^1)$$ with differential being $(f,g)\mapsto df-gd$.

Then, $\operatorname{hom}^*(C^*,C^*)\otimes C^*$ is the complex \begin{align}\operatorname{hom}(C^0,C^0)\times \operatorname{hom}(C^1,C^1)\otimes C^0\rightarrow\\ \operatorname{hom}(C^0,C^1)\otimes C^0\oplus (\operatorname{hom}(C^0,C^0)\times \operatorname{hom}(C^1,C^1))\otimes C^1\rightarrow& \operatorname{hom}(C^0,C^1)\otimes C^1\end{align} The first differential is given by $(f,g)\otimes c\mapsto (df-gd)\otimes c \oplus (f,g)\otimes dc$. The second differential is given by $u\otimes c_0\oplus (f,g)\otimes c_1\mapsto (df-gd)\otimes c_1$.

The evaluation map is the map to the complex $C^0\rightarrow C^1$ given

• in degree 0 by : $(f,g)\otimes c\mapsto f(c)$
• in degree 1 by : $u\otimes c_0 \oplus (f,g)\otimes c_1\mapsto u(c_0)+g(c_1)$

Let us check that this is indeed a morphism of complexes, in other words, let us check that these maps commute with the differentials. There are only one commutative square to check :

The first is $$\require{AMScd} \begin{CD} \operatorname{hom}(C^0,C^0)\times \operatorname{hom}(C^1,C^1)\otimes C^0@>>>\operatorname{hom}(C^0,C^1)\otimes C^0\oplus (\operatorname{hom}(C^0,C^0)\times \operatorname{hom}(C^1,C^1))\otimes C^1\\ @VVV@VVV\\ C^0@>>>C^1 \end{CD}$$ the map going through the left arrow is $(f,g)\otimes c\mapsto f(c)\mapsto d(f(c))$. While the map going through the above arrow is $$(f,g)\otimes c\mapsto (df-gd)\otimes c\oplus (f,g)\otimes dc\mapsto d(f(c))-g(dc) + g(dc)=d(f(c))$$ So the square is indeed commutative.

• Thank you. Apart from my mistake in $\hom^*(C^*, C^*)$, I think my problem arose from not knowing which morphism to apply to which element. – Bubaya Apr 13 '17 at 9:45
• @Bubaya Does my answer clear that up ? Let me know if I should explain more. For this problem, I think you should take the time to write carefully every object and every morphism from the very definition. With experience, you will be much faster at this. – Roland Apr 13 '17 at 10:18
• I think my problem was that for instance in degree 0 in $\text{ev}((f, g)\otimes c)$ I did not know what to do with the $g$ since it has domain in the wrong degree. – Bubaya Apr 13 '17 at 14:53