# Image of a category under a functor need not be a category? [duplicate]

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I've been trying to understand the following counterexample posted here: https://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics?page=2&tab=votes#tab-top

"You only need two objects 0, 1 and a single non-identity arrow 0->1. Consider the functor from this to Set which takes 0 and 1 to the natural numbers and the arrow to the successor function. "

I don't get why the image of $0\to 1$ in $Set$, which is $0\to 1$ in $\mathbb N$ is not a category...what am I missing? All the axioms seem to check out.

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• That is not the image. The image is $\text{succ} : \mathbb{N} \to \mathbb{N}$. It isn't closed under composition. – Qiaochu Yuan Jul 25 '14 at 3:51
• Why isn't it closed under composition? Since there are only two objects, there aren't any nontrivial compositions, right? – Nishant Jul 25 '14 at 4:02
• Right, in the source category there are no non-trivial compositions. But the image is the single object $\mathbb{N}$ that both objects map to. Thus, there is a composition $succ \circ succ$ – JHance Jul 25 '14 at 4:04
• Ooh, I see, I thought it was mapping to two different copies of $\mathbb N$. Thanks! – Nishant Jul 25 '14 at 4:06
• And the counterexample here: math.stackexchange.com/questions/413138/… is a counterexample because the arrow from $x\to z$, which is the composition of $x\to y$ and $y\to z$, is not in the image? – Nishant Jul 25 '14 at 4:12

Assuming we're working with small categories $\mathcal{C,D}$, the problem with images not being categories requires, for a functor $F$, the map of sets $F: Ob(\mathcal{C}) \to Ob(\mathcal{D})$ not being injective (though this isn't sufficient to insure a problem). The issue that arises is that there may be objects and morphisms $f:W \to X, g:Y \to Z$ in $\mathcal{C}$, where we have no meaningful idea of composition $f \circ g$, but such that $F(X) = F(Y)$ in $\mathcal{D}$ so that $F(f) \circ F(g)$ now makes sense. But we have no reason a priori to assume there's a map $h: X \to Z$ "lifting" the composition.
The easiest way to construct an example is to consider the finite category $\mathcal{I}$ with objects $0,1$ and a unique morphism $f: 0 \to 1$ (obviously there are also the identities). Suppose we have a functor $F: \mathcal{I} \to \mathcal{C}$, for some other category, and suppose $F(0) = F(1) = X \in Ob(\mathcal{C})$. In order for the image to be a category, $F(f) \circ F(f)$ has to be in the image, but the only morphisms in the image are $F(f)$ and the identity on $X$. So the image of $F$ is only a category if $F(f)$ is idempotent, or if it is its own inverse.
The source of the problem, as described by JHance, can be realized as the smallest 'free' example of a functor $F\colon C\to D$ whose image is not a category. Let $C$ be the category with four (distinct) objects $w$, $x$, $y$, $z$ and the two arrows $w\to x$ and $y\to z$ (besides the identities), and let $D$ be the category with the objects $0$, $1$, $2$ and the (non-identity) arrows $0\to 1\to 2$ and $0\to 3$. (Both $C$ and $D$ are partially ordered sets regarded as categories.) Let $F$ be the functor $C\to D$ with the object part $w\mapsto 0$, $x\mapsto 1$, $y\mapsto 1$, $z\mapsto 2$, which already determines the behavior of $F$ on morphisms, namely $F(w\to x)=(0\to 1)$ and $F(y\to z)=(1\to 2)$. The image of $F$ is not a category, because it contains the composable arrows $0\to 1$ and $1\to 2$, but no arrow $0\to 2$.