Lemma 12.17.3. In a Karoubian additive monoidal category every summand of an object which has a left dual has a left dual.

**Proof.**
We will use Categories, Lemma 4.43.6 without further mention. Let $X$ be an object which has a left dual $Y$. We have

If $a : X \to X$ corresponds to $b : Y \to Y$ then $b$ is the unique endomorphism of $Y$ such that precomposing by $a$ on

is the same as postcomposing by $1 \otimes b$. Hence the bijection $\mathop{\mathrm{Hom}}\nolimits (X, X) \to \mathop{\mathrm{Hom}}\nolimits (Y, Y)$, $a \mapsto b$ is an isomorphism of the opposite of the algebra $\mathop{\mathrm{Hom}}\nolimits (X, X)$ with the algebra $\mathop{\mathrm{Hom}}\nolimits (Y, Y)$. In particular, if $X = X_1 \oplus X_2$, then the corresponding projectors $e_1, e_2$ are mapped to idempotents in $\mathop{\mathrm{Hom}}\nolimits (Y, Y)$. If $Y = Y_1 \oplus Y_2$ is the corresponding direct sum decomposition of $Y$ (Section 12.4) then we see that under the bijection $\mathop{\mathrm{Hom}}\nolimits (Z' \otimes X, Z) = \mathop{\mathrm{Hom}}\nolimits (Z', Z \otimes Y)$ we have $\mathop{\mathrm{Hom}}\nolimits (Z' \otimes X_ i, Z) = \mathop{\mathrm{Hom}}\nolimits (Z', Z \otimes Y_ i)$ functorially as subgroups for $i = 1, 2$. It follows that $Y_ i$ is the left dual of $X_ i$ by the discussion in Categories, Remark 4.43.7. $\square$

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