CNN - upsampling backprop gradients across average-pooling layer How to up-sample gradients, during back-propagation, across an average-pooling layer?
For this purpose, let
$$
A^{[l]} = \begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33} \\
\end{bmatrix}; \quad P^{[l]} = \begin{bmatrix}
p_{11} & p_{12} \\
p_{21} & p_{22} \\
\end{bmatrix} = \begin{bmatrix}
\frac{a_{11} + a_{12} + a_{21} + a_{22}}{4} & \frac{a_{12} + a_{13} + a_{22} + a_{23}}{4} \\
\frac{a_{21} + a_{22} + a_{31} + a_{32}}{4} & \frac{a_{22} + a_{23} + a_{32} + a_{33}}{4} \\
\end{bmatrix} \\
\frac{\mathrm{d}J}{\mathrm{d}P^{[l]}} = \begin{bmatrix}
\frac{\mathrm{d}J}{\mathrm{d}p_{11}} & \frac{\mathrm{d}J}{\mathrm{d}p_{12}} \\
\frac{\mathrm{d}J}{\mathrm{d}p_{21}} & \frac{\mathrm{d}J}{\mathrm{d}p_{22}}  \\
\end{bmatrix}
$$
where, $A^{[l]}$ is the activation of layer-$l$ , and $P^{[l]}$ is the matrix obtained after average-pooling $A^{[l]}$ using a $2 \times 2$ pooling window, and $J$ is the cost function.
Given this, how to compute $\frac{\mathrm{d}J}{\mathrm{d}A^{[l]}}$?

*

*(Source) says, "the error is multiplied by $\frac{1}{2×2}$ and assigned to the whole pooling block (all units get this same value)." $\rightarrow$ then what would be the gradient $\frac{\mathrm{d}J}{\mathrm{d}a_{22}}$, which is a member of all the pooling blocks.

*(Source) the question mentions the up-sampling strategy, but doesn't mention what $\beta$ is.

*(Source) says, "if we have mean pooling then upsample simply uniformly distributes the error for a single pooling unit among the units which feed into it in the previous layer." $\rightarrow$ which is still vague.

*(Source) Zhang, Zhifei. "Derivation of backpropagation in convolutional neural network (cnn)." University of Tennessee, Knoxville, TN (2016). $\rightarrow$ pg.4, eq.32 results in the following up-sampled gradients
$$
\frac{\mathrm{d}J}{\mathrm{d}A^{[l]}} = \begin{bmatrix}
\frac{1}{4}\frac{\mathrm{d}J}{\mathrm{d}p_{11}} & \frac{1}{4}\frac{\mathrm{d}J}{\mathrm{d}p_{11}} & \frac{1}{4}\frac{\mathrm{d}J}{\mathrm{d}p_{12}} \\
\frac{1}{4}\frac{\mathrm{d}J}{\mathrm{d}p_{11}} & \frac{1}{4}\frac{\mathrm{d}J}{\mathrm{d}p_{11}} & \frac{1}{4}\frac{\mathrm{d}J}{\mathrm{d}p_{12}} \\
\frac{1}{4}\frac{\mathrm{d}J}{\mathrm{d}p_{21}} & \frac{1}{4}\frac{\mathrm{d}J}{\mathrm{d}p_{21}} & \frac{1}{4}\frac{\mathrm{d}J}{\mathrm{d}p_{12}}  \\
\end{bmatrix} 
$$
Is this correct? And, if it is, then is there a mathematical basis for why the gradients are up-sampled this way?

It is not right. When the pool windows overlap, derivatives must be added. This is not explicitly stated in Sources 1-3. I couldn't see it in Source 4 as well but I'm not too sure as the document is long.
Addition happens simply because of the chain rule:
$$\frac{\partial J}{\partial a_{ij}}=\sum_{m,n}\frac{\partial J}{\partial p_{mn}}\frac{\partial p_{mn}}{\partial a_{ij}}$$
For example, for $a_{12}$, we'll have
$$\begin{align}\frac{\partial J}{\partial a_{12}}&
=\frac{\partial J}{\partial p_{11}}\frac{\partial p_{11}}{\partial a_{12}}
+\frac{\partial J}{\partial p_{12}}\frac{\partial p_{12}}{\partial a_{12}}
+\frac{\partial J}{\partial p_{21}}\frac{\partial p_{21}}{\partial a_{12}}
+\frac{\partial J}{\partial p_{22}}\frac{\partial p_{22}}{\partial a_{12}}\\
&=\frac{\partial J}{\partial p_{11}}\frac{1}{4}
+\frac{\partial J}{\partial p_{12}}\frac{1}{4}
+\frac{\partial J}{\partial p_{21}}0
+\frac{\partial J}{\partial p_{22}}0\\
&=\frac{\partial J}{\partial p_{11}}\frac{1}{4}
+\frac{\partial J}{\partial p_{12}}\frac{1}{4}
\end{align}$$
It doesn't matter if this is pooling, upsampling/downsampling etc. We have a bunch of numbers $a_{ij}$, we transform and calculate a bunch of another numbers $p_{mn}$ and propagate until the loss is calculated. This should be treated like a usual transformation in order to comply with the mathematics.
