relationship between Exact sequences and Universal mapping property

Question

I am stuck in the following question.

Show that for any two short exact sequences.

$0\overset{}{\rightarrow}K\overset{i}{\rightarrow}V\overset{T}{\rightarrow}U$

and

$0\overset{}{\rightarrow}K'\overset{i'}{\rightarrow}V\overset{T}{\rightarrow}U$

then there exists a unique isomorphism $\mu:K'\rightarrow K $ satisfying $i'=i\circ\mu$

I have proved that for any short exact sequence

$0\overset{}{\rightarrow}K\overset{i}{\rightarrow}V\overset{T}{\rightarrow}U$, it satisfies the Universal Mapping Property of the kernel of $T$. I think this might be helpful to answering the problem. However, I could not see how. Does $K'$ and $K$ necessarily be isomorphic? Intuitively, $K'$ and $K$ might have different cardinality right? I am a little confused with the definition. I think I can find a surjective $\mu$ satisfying the condition, and another one in the other direction.. Will this lead to the right result? I really appreciate that someone can point to me what I am missing. Also someone recommend me a book about this whole universal mapping property theory..I am taking a linear algebra course and it's the course content, but I could not find these topics on any linear algebra book..Steve Roman covers some of the topics that we cover; however, not this part.

Answer 1

This is a standard case of "uniqueness" of final objects in some categories.

$T\circ i'=0 \Rightarrow \exists !\ \alpha \colon K'\to K $ so that $i' = i \circ \alpha$.

$T\circ i=0 \Rightarrow \exists !\ \beta \colon K\to K' $ so that $i = i' \circ \beta$.

Now we get

$i' \circ (\beta\circ \alpha) = (i' \circ \beta)\circ \alpha = i \circ \alpha = i'$ and similarly

$i \circ (\alpha \circ \beta) = (i \circ \alpha)\circ \beta = i' \circ \beta = i$

By uniqueness we conclude that $\alpha \circ \beta = \mathbb{1}_{K}$ and $\beta \circ \alpha = \mathbb{1}_{K'}$

Answer 2

Since the sequences are exact, $i$ and $i'$ are injective hence isomorphisms onto their images, thus $$ K\simeq i(K)=\ker T=i'(K')\simeq K' $$ where the equalities are due to exactness. So $K$ and $K'$ are isomorphic, to answer your question.

Anyway, I believe the following works. Let $\{y_j\}$ be a basis for $K'$. For $y_j\in K'$, $i'(y_j)\in i'(K')=i(K)$, so there exists $k_j\in K$ such that $i(k_j)=i'(y_j)$, and this $k_j$ is unique since $i$ is injective. Define $u\colon K'\to K$ by $u(y_j)=k_j$. Then $i'$ and $i\circ u$ agree on $\{y_j\}$, so $i'=i\circ u$. Then $u$ satisfies the condition, and is the unique map to do so.

To see $u$ is an isomorphism, note that $i$ and $i'$ are isomorphisms onto their images, hence map bases to bases. It follows that $\{i'(y_j)\}$ is a basis for $\ker T$. Since $i$ sends $\{k_j\}$ onto a basis for $\ker T$, $\{k_j\}$ is a basis for $K$. Then $u$ sends a basis of $K'$ to a basis of $K$, hence is an isomorphism.