## Franklin's Notes

### Injective object

In category theory , an injective object of a category $\mathsf{C}$ is an object $X\in\mathsf{C}$ with the property that, for every monomorphism $f:X\rightarrowtail Y$ and morphism $g:X\to Q$, there exists a morphism $h:Y\to Q$ such that $h\circ f = g$, or such that the following diagram commutes: The morphism $h$ can be thought of as an "extension" of $g$ from the "smaller" domain $X$ to the "larger" domain $Y$.

Why does this definition restrict $f$ to monomorphisms? Suppose that $f$ were not a monomorphism, and that $a,b:W\to X$ were such that $fa=fb$. Then it would follow that $hfa=hfb$ for any $h:Y\to Q$, making it impossible for $hf$ to be a monomorphism, so that if $g$ were a monomorphism, it would trivially be impossible for the diagram to commute. This would mean that, if the definition did not restrict $f$ to monomorphisms, no objects which are the domain of both a monomorphism and a non-monic morphism could be injective. Since most objects that we care about in nontrivial categories have both monic and non-monic morphisms originating from them, this would make the definition far too strict, ruling out most "interesting" objects.

Proposition 1. If $X_1$ is an injective object and $f_1:X_1\to X_2$ is monic and epic, then $X_2$ is an injective object.

Proof. Suppose $X_1$ is injective and $f_1:X_1\to X_2$ is monic and epic. Further let $Y,Q$ be arbitrary objects with morphisms $f_2:X_2\rightarrowtail Y$ monic and $g:X_2\to Q$. We have the following diagram, and want to show that there exists $h:Y\to Q$ such that $hf_2 = g$, so that the following diagram commutes: Notice that $f_2f_1:X_1\to Y$ is monic because $f_2$ and $f_1$ are both monic. This means that, by the injectivity of $X_1$, there exists a morphism $\alpha:Y\to X_2$ such that $\alpha f_2 f_1 = f_1$, implying $\alpha f_2 = 1_{X_2}$ by epicness of $f_1$. Then we may simply let $h=g\alpha$ to guarantee that $hf_2 = g\alpha f_2 = g$, as desired. $\blacksquare$

Corollary 1. If $X_1, X_2$ are isomorphic objects, then one is injective iff the other is also injective.

Injective objects can be equivalently defined by the following (apparently weaker) criterion, which is a special case of the above definition in which $Q=X$ and $g=1_X$.

Proposition 2. An object $X\in\mathsf{C}$ is injective iff, for all objects $Y$ and monomorphisms $f:X\to Y$ in the category, there exists $h:Y\to X$ such that the following diagram commutes: Proof. Clearly this is necessary, because this is a special case of the definition in which $Q=X$ and $g=1_X$. To see why this is sufficient, let $g:X\to Q$ be arbitrary, and let $f:X\to Y$ be monic. If there exists $h$ such that $hf=1_X$, then letting $h'=gh$ yields the equality $h'f=g$, meaning that the original definition of an injective object is satisfied. $\blacksquare$

Corollary 2. An object $X$ is injective iff every monomorphism with domain $X$ is split monic.

Next, we turn to some examples in specific categories.

Proposition 3. All objects in $\mathsf{Set}$ are injective.

Proof. Monomorphisms in $\mathsf{Set}$ are simply injections. Let $f:X\to Y$ be mono and $g:X\to Q$ be arbitrary. Then, if we let $f(X)\subset Y$ denote the image of $X$ under $f$, let $\iota:f(X)\to Y$ be the inclusion map, and let $\iota^{-1}$ be an arbitrary left-inverse of $\iota$ (since $\iota$ is injective), then we have that $\iota^{-1}f$ is a bijection, and we may let $h=g(\iota^{-1}f)^{-1}\iota^{-1}$ so that $hf=g$ as desired. Since $f,g,X$ were arbitrary, we have that all sets are injective objects in $\mathsf{Set}$. $\blacksquare$

Proposition 4. The trivial group ${e}$ is the only injective object in $\mathsf{FinGrp}$, the category of finite groups.

Proof. Suppose for the sake of contradiction that $G\ne {e}$ is an injective object of $\mathsf{FinGrp}$. All elements of $G$ have order dividing $|G|$, so that $x^{|G|}=e$ for all $x\in G$, and no non-identity element of $G$ has a $|G|$th root.

Let us construct a group $H$ with the following presentation: the generators are $y_1,...,y_{|G|}$, and the relations are given by $y_i^{|G|}y_j^{|G|}=y_k^{|G|}$ whenever $x_ix_j=x_k$ for the corresponding elements $x_i,x_j,x_k\in G$. Then we clearly have a monic homomorphism $f:G\to H$ defined by $x_i \mapsto y_i^{|G|}$ for each $x_i\in G$ and corresponding $y_i\in H$.

Let $X=H$, $Q=G$, and $g=1_G:G\to G$. We claim that there cannot exist $h:H\to G$ such that $hf = g = 1_G$. For if this were true, it would follow that $h(x_i)^{|G|}=h(x_i^{|G|})=hf(x_i)=x_i$, making $h(x_i)\in G$ a $|G|$th root of $x_i\in G$, which is impossible as remarked at the beginning. Hence $G$ is not an injective object. $\blacksquare$

The classification of injective objects in $\mathsf{LinOrd}$ is more interesting and is nontrivial unlike the above examples.

Proposition 5. In the category $\mathsf{LinOrd}$ of linear orders with monotonic functions as morphisms, the injective objects are precisely the orders which have a first element, a last element, and no gaps.

Proof. Suppose that $X$ is an ordered set with a first element, a last element, and no gaps. Let $f:X\rightarrowtail Y$ be mono (injective) and $g:X\to Q$ be an arbitrary morphism. If $y\in Y$ is not in the image $f(X)$, then there is either a largest element $f(x)\in f(X)$ less than $y$, or a smallest element $f(x)\in f(X)$ greater than $y$. For if there were not, either $y$ would define a gap of $X$, or $X$ would have no first element or no last element. We may therefore define a function $h:Y\to Q$ as follows:

1. If $y=f(x)\in f(X)$, then let $h(y)=x$.
2. Else, if there is a largest element $f(x)\in f(X)$ smaller than $y$, let $h(y)=x$.
3. Else there must exist a smallest element $f(x)\in f(X)$ greater than $y$, so we may let $h(y)=x$.

From here, it is easy to verify that $h$ preserves $\leq$ and that $hf=g$, since it fixes all elements of the image $f(X)$. Hence all linear orders with a first element, a last element, and no gaps are injective objects.

Now for the opposite direction: suppose $X$ is an ordered set with no first element, or no last element, or a gap. Then $X$ can be written as $X_1+X_2$ where $X_1$ has no last element and $X_2$ has no first element, and we may let $Y=X_1+1+X_2$ and $f$ be an inclusion map. Further let $Q=X$ and $g=1_X$. Suppose that there exists $h:Y\to Q$ such that $hf=g=1_X$, so that $h$ is a left inverse of $f$. Now let $y$ be the unique element of $Y\backslash f(X)$. Since $y$ is not in the image $f(X)$, we have that either $fh(y)>y$ or $fh(y)<y$, and in either case there exist two distinct elements $y_1,y_2 \in f(X)$ strictly between $fh(y)$ and $y$. (Checking this requires some basic casework.) However, $hfh(y)=1_Xh(y)=h(y)$, so $fh(y)$ and $y$ have the same image under the order-preserving map $h$, meaning that $h(y_1)=h(y_2)=h(y)$ by a squeezing argument. However, since $h$ is a right inverse of $f$, it must be injective on the image $f(X)$, but it sends $y_1,y_2\in f(X)$ to the same element of $X$, which is a contradiction. Hence no such $h$ can exist and $X$ is not an injective object.

We have shown that every ordered set with a first element, a last element, and no gaps is an injective object, and that no ordered set which does not meets these specifications can be an injective object, completing the proof of the proposition. $\blacksquare$

The classification of injective objects in the category $\mathsf{Met}$ (metrics spaces with distance-non-increasing functions as morphisms) is even more interesting.