Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Reading Hungerford's Algebra I encountered the following exercise:

(A) Let $\mathcal{C}$ and $\mathcal{D}$ be categories. Let $S,T: \mathcal{C} \to \mathcal{D}$ be covariant functors. If $\alpha: S \Longrightarrow T$ is a natural isomorphism, then there exists a natural isomorphism $\beta: T \Longrightarrow S$ such that $\alpha \circ \beta = \operatorname{id}_T$ and $\beta \circ \alpha = \operatorname{id}_{S}$ where $\operatorname{id}_{T,S}$ are the identity natural isomorphisms.

(B) Generalize the above statement to multifunctors (functors with product category as domain).

Note that Hungerford defines a category as being locally small. For (A), I present the following proof:

By hypothesis $\alpha_X: S(X) \to T(X)$ is an equivalence for each $X \in \operatorname{ob}(\mathcal{C})$. Hence for each component $\alpha_X$ there exists a unique morphism $\beta_X \in \hom_{\mathcal{D}}(T(X),S(X))$ such that $\alpha_X \circ \beta_X = \varepsilon_{T(X)}$ and $\beta_X \circ \alpha_X = \varepsilon_{S(X)}$.

Now define $\beta = \{\beta_X\}_{X \in \operatorname{ob}(\mathcal{C})}$. It is claimed that this is a natural isomorphism. By hypothesis $\alpha: S \Longrightarrow T$ is a natural isomorphism, and hence \begin{align*} Sf \circ \beta_X &= \varepsilon_{S(Y)} \circ Sf \circ \beta_X \\ &= (\beta_Y \circ \alpha_Y) \circ Sf \circ \beta_X\\ &= \beta_Y \circ (\alpha_Y \circ Sf) \circ \beta_X\\ &= \beta_Y \circ (Tf \circ \alpha_X) \circ \beta_X\\ &= \beta_Y \circ Tf \circ (\alpha_X \circ \beta_X)\\ &= \beta_Y \circ Tf \circ (\varepsilon_{T(X)})\\ &= \beta_Y \circ Tf \end{align*} giving commutativity; that each $\beta_X$ is an equivalence follows by hypothesis. Finally, observe that for each object $X \in \operatorname{ob}(\mathcal{C})$, \begin{equation*} (\alpha \circ \beta)_X = \alpha_X \circ \beta_X = \varepsilon_{T(X)} \implies \alpha \circ \beta = \operatorname{id}_T \end{equation*} and symmetrically $\beta \circ \alpha = \operatorname{id}_S$.

My question is why does this not hold identically for multifunctors, simply taking $\mathcal{C}$ to be the relevant product category?

share|cite|improve this question
It seems to me that B is a special case of A where we set $\mathcal{C}=\mathcal{E}^n$. – Baby Dragon Jul 26 '13 at 0:28
It does hold for multifuntors too. The author just wants you to write down the (more general) equations in this case. I would replace "equivalence" with "isomorphism" in your text and perhaps use the more common id in place of $\varepsilon$ . – magma Jul 26 '13 at 8:18
I mean strictly speaking the first statement does imply the second. I am wondering however if this is just an authorial oversight or if there is some sort of pedagogical content I could be missing involved in considering the special case. – Cameron Jul 26 '13 at 23:25
Cameron, you are right, B is a special case of A. But Hungerford already doesn't gives the best definition of a natural isomorphism (actually A is the correct one, and what Hungerford gives, should be called a natural transformation which is a pointwise isomorphism, which turns out to be a characterization of natural isomorphisms). Many books ignore forgetful functors as if they don't play any role ... – Martin Brandenburg Aug 5 '13 at 18:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.