How the map in proving the direct limit property $\lim\limits_{\to}(M_i\times N)\cong(\lim\limits_{\to}M_i)\times N$ is well defined

Question

I want to show the direct limit of $(M_i\times N,\mu_{ij}\times id_N)$ is $M\times N$ where $M$ is the direct limit of $(M_i,\mu_{ij})$.

Suppose $Q$ be any $R-$ module. Let $\alpha_i:M_i\times N\to Q$ be a collection of $R-$linear maps. Then for all $i\leq j$ we have $$\alpha_i=\alpha_j\circ (\mu_{ij}\times id_N)$$ Now we want to define a unique map $\alpha:M\times N \to Q$ such that $\alpha_i=\alpha\circ (\mu_i\times id_N)$.

Now every element of $M$ is of the form $\mu_i(x_i)$ for some $i\in I$ where $x_i\in M_i$. So we define $$\alpha(x,y)=\alpha(\mu_i(x_i),y)=\alpha_i(x_i,y)$$

Now I need to show that $\alpha$ is well defined. This is where I am getting problems. Suppose $x=\mu_i(x_i)=\mu_j(x_j)$. Since $I$ is direct set, $\exists\ k\in I$ such that $k\geq i,j$. So $$\mu_i(x_i)=\mu_k\circ \mu_{ik}(x_i)=\mu_k\circ\mu_{jk}(x_j)=\mu_j(x_j)$$ since $\mu_i=\mu_k\circ \mu_{ik}$ and $\mu_j=\mu_k\circ\mu_{jk}$. Then $$\alpha(\mu_i(x_i),y)=\alpha_i(x_i,y)=\alpha_k(\mu_{ik}(x_i),y)$$and $$\alpha(\mu_j(x_j),y)=\alpha_j(x_j,y)=\alpha_k(\mu_{jk}(x_j),y)$$Now how to show that these two are equal

Answer 1

You can prove this general statement first

If $x_k\in M_k$ is such that $\mu_kx_k=0$, then there is some $l\geq k$ such that $\mu_{kl}x_k=0$ in $M_l$.

Then, since $\mu_k(\mu_{ik}x_i-\mu_{jk}x_j)=0$, there is some $l\geq k$ such that $$\mu_{kl}(\mu_{ik}x_i-\mu_{jk}x_j)=0 \in M_l.$$ I.e. $\mu_{il}x_i=\mu_{jl}x_j$. Hence $$\alpha_k(\mu_{ik}x_i,y)=\alpha_l(\mu_{il}x_i,y)=\alpha_l(\mu_{jl}x_j,y)=\alpha_k(\mu_{jk}x_j,y)$$ as you wanted.

Answer 2

Well, you’ve missed out on a hypothesis or two. It’s not true that for any family of module homomorphisms $\alpha_\bullet:M_\bullet\times N\to Q$ there is one and only one extension $\alpha:M\times N\to Q$: you assume also the $\alpha_\bullet$ are compatible with the diagram (form a cocone under it) in the sense that for all $i$ and $j$, $\alpha_j\circ(\mu_{ij}\times1_N)=\alpha_i$. You didn’t quite elaborate on this but I’m assuming $\mu_{ij}:M_i\to M_j$.

It is a necessary assumption that the diagram of $(\mu_{ij})_{i,j\in I}$ has only one connected component.

Split each $\alpha_i$ into components $\beta_i:M_i\to Q$ and $\gamma_i:N\to Q$ (using $A\times B\cong A\oplus B$ for RMod). We wish to show there is a unique $\alpha=\beta+\gamma:M\oplus N\to Q$ with $\beta\mu_\bullet\equiv\beta_\bullet$ and $\gamma1_N\equiv\gamma_\bullet$. That's the same thing as showing that $\beta,\gamma$ exist uniquely (individually).

Firstly notice that the condition on $\alpha_\bullet$ ensures $\gamma_j=\gamma_j\circ1_N=\gamma_{i_1}=\cdots=\gamma_{i_N}=\gamma_i$ for all $i,j$ and some $i_1,i_2,\cdots,i_N$ that connect $i$ to $j$ in the diagram. It follows all $\gamma_i$ equal all other $\gamma_j$. Instantiate a particular $j_0\in I$ and define $\gamma:=\gamma_{j_0}$. It's then clear $\gamma$ exists uniquely.

The "necessary assumption" is necessary because if there existed $i,j$ with $\gamma_i\neq\gamma_j$ (which is possible iff. the diagram is not connected), such a $\gamma$ could not possibly exist. Consider $(M_1\oplus N)\oplus(M_2\oplus N)\not\cong(M_1\oplus M_2)\oplus N$ in general (this is the picture if no $\mu_{ij}$ exist at all).

What about $\beta$? Well, we seek a $\beta$ with $\beta\mu_i=\beta_i$ for all $i$. Such a homomorphism $\beta$ exists uniquely precisely by definition of $M$ as $\varinjlim(M_i,\mu_{ij})_{i,j\in I}$.