Snake lemma without elements – exactness


As mentioned in this answer, the connecting homomorphism $$\ker l\to\coker m$$ is constructed as follows:

• $$lgk_lg=0$$, so by the universal property of $$\ker l$$, the map $$gk_{lg}$$ factors through a morphism $$\tilde g\colon\ker lg\to\ker l$$.
• $$g$$ is epic. Applying left exactness of the kernel sequence (which we assume to be already shown) shows that $$\tilde g$$ and hence $$\xi$$ is epic. Question: why is $$\xi$$ an isomorphism?
• $$k_l\tilde g\alpha=gf=0$$, so by the universal property of $$\coker\alpha$$, the map $$\tilde g$$ factors through a morphism $$\xi\colon\coker\alpha\to\ker l$$.
• $$lgf=0$$, so by the universal property of $$\ker lg$$, the map $$lg$$ factors through a morphism $$\alpha:M\to\ker lg$$.
• $$g'nk_{lg}=lgk_{lg}=0$$, so by the universal property of $$\ker g'$$, the map $$g'n$$ factors through a morphism $$\beta\colon\ker lg\to\ker g'=\im f'=M'$$, where the latter equality holds because $$f'$$ is monic.
• $$c_m\beta\alpha=c_mm=0$$, so by the universal property of $$\coker\alpha$$, the map $$c_m\beta$$ factors through a map $$\gamma\colon\coker\alpha\to\coker m$$. The connecting homomorphism then is $$\delta=\gamma\xi^{-1}$$.

Now we want to show exactness at $$\ker l$$. From

One sees that:

• $$lgk_n=g'nk_n=0$$, so by the universal property of $$\lg$$, the map $$\ker n\to N$$ factors through a morphism $$k_n\colon\ker n\to\ker lg$$.
• $$f'\beta k_n=nk_n=0$$, so since $$f'$$ is monic, $$\beta k_n=0$$. Then $$\delta \tilde gk_n=c_m\beta k_n=0$$, so bt the universal property of $$\ker\delta$$, the map $$\tilde gk_n$$ factors through $$\tilde{\tilde{g}}\colon\ker n\to\ker\delta$$. This shows the one inclusion.

Question: I have no idea how to show surjectivity onto $$\ker\delta$$ of this factorisation. How do I do that?

Notations: If $$f:X\to Y$$ and $$g:Y\to Z$$, then $$fg:X\to Z$$ denote composition.
I write $$f\propto g$$ if there exists a morphism $$h$$ such that $$f=gh$$, while the reversed symbol of $$\propto$$ if there exists a morphism $$h$$ such that $$f=hg$$. If $$f=hg$$ with $$g$$ monic, then $$h$$ is uniquely determined and denoted with $$\frac fg$$. If $$f=hg$$ with $$h$$ an isomorphism, then I write $$f\simeq g$$.
If $$f:X\to Z$$ and $$g:Y\to Z$$ then $$f\mathrel\urcorner g$$ and $$f\mathrel\ulcorner g$$ denote pullback projections, so that $$(f\mathrel\urcorner g)f=(f\mathrel\ulcorner g)g$$.
If $$f:Z\to X$$ and $$G:Z\to Y$$ then $$[f,g\rangle$$ denote the morphism $$Z\to X\times Y$$ into the product. In particular, I write $$f\mathrel\top g=[f\mathrel\urcorner g,f\mathrel\ulcorner g\rangle$$.